Ultraproducts

David Pierce

August , 

Mimar Sinan Güzel Sanatlar Üniversitesi Matematik Bölümü http://mat.msgsu.edu.tr/~dpierce/ Ultraproducts

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Mathematics Department Mimar Sinan Fine Arts University Istanbul, Turkey http://mat.msgsu.edu.tr/~dpierce/ [email protected]  Preface

These notes are for a course called Ultraproducts and Their Conse- quences, to be given at the Nesin Mathematics Village in Şirince, Selçuk, İzmir, Turkey, in August, . The notes are mainly for my use; they do not constitute a textbook, although parts of them may have been written in textbook style. The notes have not been thoroughly checked for correctness; writing the notes has been my own way of learning some topics. The notes have grown like a balloon, at all points: I have added things here and there as I have seen that they are needed or useful. I have also rearranged sections. There is too much material here for a week- long course. Some of the material is background necessary for thorough consideration of some topics; this background may be covered in a simul- taneous course in Şirince. The catalogue listing for the course(with abstract as submitted by me on January , ) is as follows. Title of course: Ultraproducts and their consequences Instructor: Assoc. Prof. David Pierce Institution: Mimar Sinan GSÜ Dates: – Ağustos  Prerequisites: Some knowledge of algebra, including the theorem that a quotient of a ring by an ideal is a field if and only if the ideal is maximal. Level: Advanced undergraduate and graduate Abstract: An ultraproduct is a kind of average of infinitely many struc- tures. The construction is usually traced to a  paper of Jerzy Los; however, the idea of an ultraproduct can be found in Kurt

From http://matematikkoyu.org/etkinlikler/2012-tmd-lisans-lisansustu/ ultra_pierce.pdf, to which there is a link on http://matematikkoyu.org/ etkinlikler/2012-tmd-lisans-lisansustu/ as of August , .

  Preface

Goedel’s  proof (from his doctoral dissertation) of the Com- pleteness Theorem for first-order logic. Non-standard analysis, de- veloped in the s by Abraham Robinson, can be seen as taking place in an ultraproduct of the ordered field of real numbers: more precisely, in an ultrapower. Indeed, for each integer, the ‘average’ real number is greater than that integer; therefore an ultrapower of the ordered field of real numbers is an ordered field with infinite elements and therefore infinitesimal elements. Perhaps the first textbook of model theory is Bell and Slomson’s Models and Ultra- products of : the title suggests the usefulness of ultraproducts in the development various model-theoretic ideas. Our course will investigate ultraproducts, starting from one of the simplest inter- esting examples: the quotient of the cartesian product of an infinite collection of fields by a maximal ideal that has nontrivial projection onto each coordinate. No particular knowledge of logic is assumed. Such was the abstract that I submitted in January. I have written the following notes since then, by way of working out for myself some of the ideas that might be presented in the course. I have tried to emphasize examples. In some cases, I may have sacrificed generality for concrete- ness. A theorem that I might have covered, but have not, is the theorem of Keisler and Shelah that elementary equivalence is the same thing as isomorphism of ultrapowers.

 Contents

 Preface 

 Notation 

 Introduction  . Productsoffields......  . Ultrapowers of the field of real numbers ......  . Orderedrings ......  . Non-standardanalysis ...... 

 Model theory  . Theoriesandmodels ......  . Definablerelations ......  . Substructures ...... 

 Ultraproducts and Łoś’s Theorem  . Idealsandfilters ......  . Reducedproducts ......  . Ultraproducts......  . Cardinality ...... 

 Simple applications  . Arrow’sTheorem......  . Compactness ......  . Elementaryclasses ......  . A countable non-standard model of arithmetic ...... 

 Gödel’s Completeness Theorem  . Formalproofs......  . Completeness by ultraproducts ......  . Completeness by König’s Lemma ......  . Arbitrary formulas ...... 

 Contents

 Boolean rings and Stone spaces  . Booleanrings ......  . Ultrafilters ......  . Stonespaces......  . Booleanoperations ...... 

 More model theory  . The structure of definable relations ......  . Lindenbaum–Tarski algebras ......  . Theoriesandtype-spaces ......  . Saturation...... 

 Rings  . Ideals ......  . Localizations ......  . Algebraicgeometry......  . AGaloiscorrespondence...... 

 Finite fields  . Ultraproducts of finite structures ......  . Finitefields ......  . Galoisgroups ......  . Pseudo-finitefields ...... 

 Schemes  . The spectrum of a polynomial ring ......  . Regularfunctions......  . Generic points and irreducibility ......  .Affineschemes ......  . Regular rings (in the sense of von Neumann) ......  . Theultraproductscheme ...... 

Bibliography 

 List of Figures

. Thediagonalmap ......  ω . An infinite element of R /M ......  . The positive cones of R2 and C ...... 

. Symmetric differences of two sets and three sets ......  . Distribution in Boolean rings ......  . AnidealofaBooleanring......  . AfilterofaBooleanring ......  . Notation for sequences of tuples ...... 

. An election with three candidates ...... 

. Acompletebinarytree...... 

. Booleancombinations ...... 

. The universal property of the quotient field ......  . The zero-locus of y x2 in R ......  − . The zero-locus of y x2,y x in R ......  . Algebraic-geometric{ Galois− correspondence− } ......  . Jointembeddingpropertyoffields ...... 

. The lattice of finite fields of characteristic p ......  . The universal property of Gal(Fp) ......  . Isomorphismsofpseudo-finitefields...... 

. Astalkofasheaf......  . The universal property of Rp ...... 

  Notation

The set-theorist’s natural numbers compose the set ω:

ω = 0, 1, 2,... . { } In these notes, this set is used:

) as an index-set for countably infinite sequences (ak : k ω); ) as the cardinal number of each countably infinite set. ∈ The set-theoretic feature of ω is that if n ω, then ∈ n = 0,...,n 1 . { − } In particular

0= ∅, 1= 0 , 2= 0, 1 . { } { } If A and B are sets, we let

AB = functions from B to A . { } In particular, if n ω, then ∈ An = functions from n to A . { } An element of An may be written as either of

0 n−1 (b0,...,bn−1), (b ,...,b ), and this can be abbreviated by

b,

The ω here is printed in an upright, ‘roman’ font, since it has a constant meaning. This means the sloping, ‘italic’ ω is available for use as a variable; but in fact it will not be used here.

 in boldface: it is an n-tuple of elements of A. The superscripts in (b0,...,bn−1) are not exponents, but just (upper) indices; we may use them, because we shall occasionally use lower indices at the same time, n as for example in consideration of sequences (bk : k ω), where bk A , so that ∈ ∈ 0 n−1 bk =(bk,...,bk ). Note that A0 = 0 = 1. { } The ring of (rational) integers is Z, which is a sub-ring of Q, the field of rational numbers; this in turn is a subfield of R, the field of real numbers. The set of positive integers can be denoted by N. Thus

N = 1, 2, 3,... . { } One could write this also as ωr 0 . Some people also put 0 in N, and that is fine. However, in these notes,{ I} attempt to distinguish notationally the two roles of natural numbers: () as indices and () as rational numbers. In the former role, the natural numbers compose ω; in the latter, N. It is also just useful to have distinct simple symbols for the two sets 0, 1, 2,... and 1, 2, 3,... . { } { }

  Introduction

. Products of fields

Suppose K is an indexed family (K0,K1,K2,... ) or (Ki : i ω), where  ∈ each Ki is a field. For example, each Ki might be R, or each Ki might be a different finite field. We can form the Cartesian product of the family K . This product is denoted by one of the expressions

Ki, K . i∈ω Y Y If a belongs to this product, this means

a =(a : i ω), i ∈ where a K in each case. Here a is simply a function on ω that, at i ∈ i every element i of this domain, takes a value in Ki; we write this value  i as ai, though it could be written also as a(i) or a . The product K is a ring in the obvious way, with respect to the termwise operations: Q a + b =(a + b : i ω), i i ∈ a =( a : i ω), − − i ∈ 0=(0: i ω)=(0, 0, 0,... ), ∈ a b =(a b : i ω), · i · i ∈ 1=(1: i ω)=(1, 1, 1,... ). ∈ The basic of rings and fields are reviewed, for completeness, in Chapter ; but it is assumed that the reader is already somewhat familiar with them. In the present section, we could work with an arbitrary index set Ω in place of ω. Later (§.) we shall do this; but there is no obvious need to do so now. Even the notation ia might be used. Indeed, aσ is used below (p. , §.) for the image under an automorphism σ of an element a of a given field.

 . Ultrapowers of the field of real numbers

Suppose M is a maximal ideal of the ring K . Then the quotient K /M is a field, called an ultraproduct of the family K . We want to understand this ultraproduct. Q Q There is a trivial case. There may be some j in ω such that

M = x K : x = 0 . { ∈ j } Then M is a principal ideal: it isY generated by an element (1,..., 1, 0, 1,... ), where every entry is 1, except the entry 0 with index j. For, if this element of M is c, then for every a in M we have a = a c. In this case · x + M = y + M x = y . ⇐⇒ j j K Thus /M ∼= Kj under the map x xj. This isomorphism consti- tutes a proof that the ideal (c) generated7→ by c is indeed maximal (since the quotientQ of a ring by an ideal is a field if and only if the ideal is maximal). By contrast, if I is an ideal generated by an element (1,..., 1, 0, 1,..., 1, 0, 1,... ), K with two zero entries, indexed by j and ℓ respectively, then /I ∼= Kj Kℓ. Since then Kj Kℓ is not a field—since for example the nonzero elements× (1, 0) and (0, 1)×are not invertible—, I must not be maximal.Q If M is a nonprincipal maximal ideal of K , then the ultraproduct K /M will be seen to be a kind of ‘average’ of the fields Kj . But this must be properly understood. More precisely,Q if a K , then a + M Q ∈ will be a kind of average of the individual elements ai of the factors Ki. We consider a particular example in the next section.Q

. Ultrapowers of the field of real numbers

ω If each Ki is R, then the product i∈ω Ki is the power R . If M ω is a maximal ideal of this ring, then the quotient R /M is called an Q This is written out and proved below as Theorem  on page ; but the reader should already be familiar with it.

  Introduction

ω ultrapower of R. The field R embeds in R /M under the diagonal map, x (x,x,x,... )+ M. 7→ The diagonal map is so called, presumably because of the similarity of its definition to that of the map x (x,x) from R to R2, whose range is called a ‘diagonal’ line. Perhaps7→ we are likely to consider points (x,x,x,... ) of Rω themselves as horizontal lines, as in Figure .. It

x r ✲ r r r r r

0 1 2 3 4

Figure .: The diagonal map

ω should be clear that the diagonal map is an embedding of R in R /M: it is a monomorphism, with trivial kernel. For, if x = 0, then for every ω 6 element a of R we have a a a a a = 0 , 1 , 2 ,... (x,x,x,... )= i : i ω (x: i ω). x x x · x ∈ · ∈     Thus, if it were possible that x = 0, but (x,x,x,... ) M, then M would ω 6 ∈ be the improper ideal R . ω If M is a non-principal maximal ideal of R , then the diagonal embedding ω of R in R /M is not surjective: it is not an epimorphism. For, the element (1, 2, 3,... )+M of the quotient is not in its range. See Figure .. Indeed, suppose if possible

(1, 2, 3,... )+ M =(x,x,x,... )+ M,

Etymologically speaking, a diagonal line is a line joining two angles of a polygon.

 . Ultrapowers of the field of real numbers

4 r

3 r

2 r

1 r

0 1 2 3

ω Figure .: An infinite element of R /M

ω so that (1 x, 2 x, 3 x,... ) M. If x / N, then for every a in R we have − − − ∈ ∈ a a a a = 0 , 1 , 2 ,... (1 x, 2 x, 3 x,... ), 1 x 2 x 3 x · − − −  − − −  which must be in M; and this contradicts that M is proper. If x N, ∈ so that 1 x is a non-positive element n of Z, then M contains (n, n + 1,..., 1,−0, 1, 2,... ), and therefore M also contains (1,..., 1, 0, 1,... ), because− this is 1 1 1 1 1 (n, n + 1,..., 1, 0, 1, 2,... ) , ,..., , 0, , ,... . − · n n + 1 1 1 2  −  Thus M includes a principal maximal ideal as before, so M itself is either principal or improper. ω On the assumption that M is a nonprincipal maximal ideal of R , we ω have shown that the element (1, 2, 3,... )+ M of the quotient R /M is not in the image of R under the diagonal map. But this element will be seen as a kind of average of the elements of N. Indeed, for every m in N, all but finitely many elements of N are greater than m. So the ‘average’ element of N is greater than m. And we shall see that (1, 2, 3,... )+ M > (m,m,m,... )+ M. Since this will be true for all m in N, the element (1, 2, 3,... )+ M of ω R /M will be infinite.

  Introduction

ω But so far, all we know is that R /M is a field. We now want to show that it has a (linear) ordering that makes it into an ordered field.

. Ordered rings

In these notes, a ring is always a commutative unital ring. Then an or- dered ring (or more precisely a partially ordered ring) is a pair (R,R+), where R is a ring in our sense, and R+ is a subset of R that is closed under addition and multiplication and that, for all x in R, contains both x and x if and only if x = 0. That is, − x R & y R = x + y R & xy R, ∈ ∈ ⇒ ∈ ∈ x R & x R = x = 0, ∈ − ∈ ⇒ 0 R. ∈ In this case, we can define

x 6 y y x R+. ⇐⇒ − ∈ Then R+ = x R: 0 6 x . { ∈ } The set R+ is the positive cone of 6. However, only the elements of R+ r 0 should be called positive. It follows that { } () the relation 6 is reflexive, that is,

x 6 x,

since x x R+; − ∈ The Wikipedia article ‘Ordered ring’ defines this as what, for present purposes, should be called a linearly or totally ordered ring: it meets the additional condition that at least one of x and −x is always in R+. Then what for us is an ordered ring is what, for the Wikipedia article, should be called a partially ordered ring; but the article defines no such thing. At least this is the situation as of July , . The notation R+ and the term positive cone are used only in another Wikipedia article, ‘Partially ordered group’. The notation and term are unfortunate, since the positive cone in the present sense contains 0, but the real number 0 is not normally considered positive. We could define ordered rings in terms of the set {x: x> 0}; but then the possibility of positive zero-divisors would have to be dealt with; also, there would be a complication in the definition of the product order.

 . Ordered rings

() the relation 6 is anti-symmetric, that is,

x 6 y & y 6 x = x = y, ⇒ since if y x and x y are in R+, then they must be 0; () the relation− 6 is transitive,− that is,

x 6 y & y 6 z = x 6 z, ⇒ since if y x and z y are in R+, so is their sum, z x. − − − The relation 6 is thus a (partial) ordering of R, induced by R+. More- over,

() this ordering is translation-invariant, that is,

x 6 y = x + z 6 y + z, ⇒ since if y x is in R+, so is (y + z) (x + z). Finally, () multiplication− by positive elements preserves− the ordering, that is,

x 6 y & 0

+ Suppose ((Ri,Ri ): i I) is an indexed family of ordered rings. Then easily the product ∈ + Ri, Ri Yi∈I Yi∈I  is an ordered ring. For example, the ‘Cartesian plane’ R2 (that is, R R) is an ordered ring whose positive cone is the first quadrant. See Figure× .. However, this quadrant is not the positive cone of an ordering of C, since it is not closed under complex multiplication. The complex field does however have an ordering, whose positive cone consists of the non- negative real numbers (as in Figure .). In this ordering, a + bi 6 x + yi

  Introduction

✻ ✻

✲ ✲

Figure .: The positive cones of R2 and C

ω if and only if a 6 x and b = y. The product-ordering of the power R is given by the rule

0 6 x 0 6 x & 0 6 x & 0 6 x & ... 0 6 x . ⇐⇒ 0 1 2 ⇐⇒ i i∈ω ^

Suppose again (R,R+) is an arbitrary ordered ring, and I is an ideal of R. We let R+/I = x + I R/I : x I . { ∈ ∈ } It is possible that x + I R+/I although x / R+. The set R+/I is closed under addition and∈ multiplication. So it∈ is the positive cone of an ordering of R/I if and only if

x + I R+/I & x + I R+/I = x + I = I, ∈ − ∈ ⇒ that is (since x + I R+/I means x + y I for some y in R+), − ∈ ∈ x R+ & y R+ & x + y I = x I. ∈ ∈ ∈ ⇒ ∈ (By symmetry, y I can also be concluded.) For example, the only ∈ proper, non-trivial ideals of the product ring R2 are the principal ideals generated respectively by (1, 0) and (0, 1). These ideals meet the desired condition with respect to the product ordering, and the quotient (in either case) is an ordered ring that is order-isomorphic to R.

 . Ordered rings

ω ω The same is true for ideals I of R . Indeed, suppose x and y in R are positive or zero (that is, xi > 0 and yi > 0 in each case), and x + y I. Since ∈ x + y =0 = x = 0, i i ⇒ i it follows that x I. Indeed, under the hypothesis, x =(x+y)z, where ∈

xi/(xi + yi), if xi + yi = 0, zi = 6 (0, if xi + yi = 0.

ω Thus R /I is ordered. ω Now let M be a maximal ideal of R . We want to show that the ordering ω ω of R /M is linear. For every a in R , we have a + M > 0 if and only if there is b in M such that a + b > 0, that is, ai + bi > 0 in each case. If there is such an element b of M, it meets the condition

b =0 = a > 0. i ⇒ i In this case, replacing b with an appropriate product bc, we may assume

b = 0 a > 0. ( ) i ⇐⇒ i ∗ Moreover, this condition on an element b of M is sufficient—not to ensure that a + b > 0, but to ensure that a + d > 0 for some d in M. This d could be e b, where, if b = 0, so that a < 0, then · i 6 i 2ai ei = − . bi We can see now that the following are equivalent:

a + M > 0. • M has an element b such that ( ) holds. • ∗ω M contains every element b of R such that ( ) holds. • ∗ ω Suppose one of these conditions fails. Then there is some b in R r M such that ( ) holds. In this case, since M is maximal, ∗ ω M +(b)= R .

  Introduction

In particular, (1, 1, 1,... ) M +(b). Hence M has an element c such that ∈ b =0 = c = 0, i ⇒ i 6 that is, a > 0 = c = 0. i ⇒ i 6 As before, by replacing c with a multiple of it, so that ci = 0 when ai < 0, we may assume a > 0 c = 0; i ⇐⇒ i 6 ω indeed, M contains every such element c of R . Then not only does M contain c such that a + c = 0 for all i in ω, but it contains c such that i i 6 ai + ci < 0 for all i in ω. Consequently a + M < 0, under the assumption ω that a + M is not positive. Thus the ordering of R /M is linear.

. Non-standard analysis

ω Suppose now M is a non-principal maximal ideal of R . If m N, we want to show finally ∈

(1, 2, 3,... )+ M > (m,m,m,... )+ M, that is,

(1 m, 2 m, 3 m,..., 1, 0, 1, 2, 3,... )+ M > 0. − − − − It is enough to show that M has an element a such that

i

 . Non-standard analysis

We may assume that Q is included in every ordered field. An element of an ordered field that is greater than every element of Q is infinite; an element whose absolute value is less than every positive element of ω Q is infinitesimal. We have now shown that R /M has infinite ele- ments. Therefore it also has nonzero infinitesimal elements (namely the reciprocals of the infinite elements). Thus the possibility is opened up of saying that a function f from R to R is continuous at a if f(a) f(x) is infinitesimal whenever a x is infinitesimal. − − However, R itself contains no infinitesimals; so if a R, but a x is ω∈ − infinitesimal, then x / R. However, x might be in R /M. So we want ∈ ω to extend f to a function ∗f on R /M. It is obvious how to do this: If ω a R , then we should define ∈ ∗f(a + M)=(f(a ): i ω)+ M. i ∈ However, we must check that such a function ∗f exists: we must show a + M = b + M = (f(a ): i ω)+ M =(f(b ): i ω)+ M, ⇒ i ∈ i ∈ or rather a b M = (f(a ) f(b ): i ω) M. − ∈ ⇒ i − i ∈ ∈ But this is true since a b =0 = f(a ) f(b ) = 0, i − i ⇒ i − i so that, as before, (f(a ) f(b ): i ω) is a multiple of a b. i − i ∈ − We have been using implicitly (or proving special cases of) the lemma that if a =0 = b = 0 i ⇒ i ω and a M, then b M. For an arbitrary element a of R , we define ∈ ∈ supp(a)= i ω: a = 0 . { ∈ i 6 } Let us now write m for supp[M]; that is, m = supp(x): x M . { ∈ } Then M is determined by m, and moreover ω M = x R : supp(x) m . { ∈ ∈ }

  Introduction

The subset m of P(ω) has the following properties:

∅ m, ∈ X Y & Y m = X m, ⊆ ∈ ⇒ ∈ X m & Y m = X Y m. ∈ ∈ ⇒ ∪ ∈ Indeed, the first of these properties follows because 0 M. The second is by the lemma just mentioned. For the third, if supp(∈ a) = X and supp(b)= Y , then supp(c b)= Y r X for some c, and then · supp(a + c b)= X Y. · ∪

Let us write u for P(ω) r m. Then we have

ω u, ∈ X Y & X u = Y u, ⊆ ∈ ⇒ ∈ X u & Y u = X Y u. ∈ ∈ ⇒ ∩ ∈ Finally, X m X / u. ∈ ⇐⇒ ∈ Elements of m will be called small; elements of u, large. If a+M = b+M, that is, a b M, this means supp(a b) m, that is, a and b differ on a small subset− ∈ of ω, but they agree on a− large∈ subset of ω. Then the same is true of (f(ai): i ω) and (g(ai): i ω). Thus there is a well-defined function ∗f as above.∈ ∈ ω Much more now follows. For example, R /M is real-closed, that is, it satisfies the Intermediate Value Theorem for polynomial functions. In- deed, suppose p is a polynomial (in one variable) with coefficients from ω ω R /M such that, for some a and b in R /M, we have p(a) < 0 and p(b) > 0. We may assume a

i: a

The value p(a) of p at a can be written as (pi(ai): i ω), where pi is obtained from p by replacing its coefficients with their∈ values at i. Then the sets i: pi(ai) < 0 and i: pi(bi) > 0 are large, and therefore their intersection{ is large, and} therefore{ (by the} usual Intermediate Value

 . Non-standard analysis

Theorem for R) there are ci between ai and bi for a large set of i such that pi(ci) = 0. Choosing ci for the other i arbitrarily, we have a

  Model theory

. Theories and models

If n ω, then on a set A, an n-ary relation on A is just a subset of An; an n∈-ary operation on A is a function from An to A. Since A0 = 0 , so that a function on A0 takes only one value, a 0-ary or nullary operation{ } on A can be considered as an element of A. Much of mathematics is a study of operations and relations on sets. Group-theory is a study of binary operations with certain properties; field-theory, two binary operations; lattice-theory, a binary relation. A particular nonempty set, considered together with certain operations and relations on it, is called a structure. If the set is A, B, or C, then the structure might be denoted respectively by A, B, or C. The distinguished operations and relations of A are denoted by certain symbols, which constitute the signature of A. The universe of A is just the set A. More than one structure can share the same signature. For example, the signature of fields is 0, 1, , +, ; of ordered fields, 0, 1, , +, ,< (or 0, 1, , +, , 6 ). In{ general,− if·}S is a signature, then{ the− class· of} all structures{ − with· } this signature can be denoted by

Mod(S ).

If A Mod(S ), and s S , then, to avoid ambiguity, the operation or relation∈ on A that is denoted∈ in A by s can also be denoted by

sA. This is (nearly?) all that I shall say about examples in this chapter, despite the intention stated in the Preface to emphasize examples. Perhaps model-theory as a branch of mathematics can be characterized as not being about particular examples, but being an abstraction from them. In any case, the subject of these notes is not model-theory as such, but ultraproducts. It simplifies things to require the set to be nonempty, although one can certainly define empty structures if one wishes, and sometimes this is useful.

 . Theories and models

A symbol in a signature is more precisely

an n-ary predicate, if it denotes an n-ary relation, or • an n-ary operation-symbol. • The symbol for a nullary operation (that is, element) can be called a constant. The same symbol will never denote, say, an element of one set, but a binary relation on another set.

If S is a signature, then the terms of S are built up from the operation- symbols in S , along with (individual) variables. The formulas of S are then built up from the terms and predicates of S , along with log- ical symbols such as , , and (and brackets, if needed). A term is merely a recipe for obtaining¬ ∧ new∃ operations through composition of some given operations. (The coordinate projections are always understood as given.) Then a formula is a recipe for obtaining new relations through set-theoretic manipulations of given operations and relations. We shall review the formal definition of terms and formulas in the next section.

An occurrence of a variable in a formula of S will be either free or bound. If precisely the variables x0, ..., xn−1 occur freely in a formula ϕ of S , then ϕ can be written as ϕ(x0,...,xn−1) or ϕ(x). If A Mod(S ), and a An, then ϕ(a) is the result of replacing each free occurrence∈ of x in ∈ i ϕ with ai, for each i in n. This new formula ϕ(a) may not be a formula of S ; but it is a formula of S (A), which is just S along with a constant for each element of A. Such constants are often called parameters. The new formula ϕ(a) is a sentence, because it has no free variables. This sentence will be either true or false in A; if it is true, then we shall write

A  ϕ(a).

The symbol  thus stands for a relation in a more general sense: it is not a binary relation on a set, but it is a relation between certain structures and certain sentences. We might call this the satisfaction relation. (However, if A  ϕ(a), it might be said that a satisfies ϕ in A.)

The set of all sentences of S can be denoted by

Sn(S ).

  Model theory

If Γ Sn(S ), and every sentence in Γ is true in A, then we write ⊆ A  Γ; in this case, A is a model of Γ. We can define

Mod(Γ) = A Mod(S ): A  Γ = A Mod(S ): A  σ ; ( ) { ∈ } { ∈ } ∗ σ\∈Γ a class of structures of this form is called an elementary class. If Mod(S ), then we define K⊆ Th( )= σ Sn(S ): A  σ ; ( ) K { ∈ } † A \∈K a set of sentences of this form is called a theory. Both theories and elementary classes are defined in the same way, as intersections of sets defined in terms of the satisfaction relation . Therefore it turns out that there is a one-to-one correspondence between elementary classes and theories. This is a kind of Galois correspondence; see §. (and before that, ( ) on page ). † If = A , then Th( ) can be written as K { } K Th(A). Such a theory is a complete theory, namely a theory that, for every sentence σ if its signature, contains either σ or its negation σ, but not both. For an arbitrary subclass of Mod(S ), if Th( ) contains¬ both σ and σ, then must be empty,K so K ¬ K Th( ) = Sn(S ). K Suppose now Th( ) is complete. Then contains some A, and then, since K K Th( ) Th(A) Sn(S ), K ⊆ ⊂ the first inclusion must be an equation. Thus the complete theories are precisely the theories of individual structures. However, there are proper classes of structures whose theories are complete. For example, the the- ory of algebraically closed fields of characteristic 0 is complete (Theorem  on page ).

That is, classes that are not sets, because they are ‘too large’.

 . Definable relations

If Th(A) = Th(B), then A and B are said to be elementarily equiva- lent, and we write A B. ≡ If A and B are isomorphic in the familiar sense, then (trivially) they are elementarily equivalent. However, by what we have just observed about complete theories, the converse fails: non-isomorphic structures can be elementarily equivalent. This is easily shown with the Compactness The- orem, which we shall prove as Theorem  by using ultraproducts. Also, a non-principal ultrapower of an infinite structure is elementarily equiv- alent, but not isomorphic, to that structure. See also §..

. Definable relations

We are usually interested in Mod(T ) for particular theories T of a sig- nature S . One way to study this is to study the definable relations in elements of Mod(T ). Suppose A is one of these elements, and ϕ is an n-ary formula of S . Then ϕ defines the subset

a An : A  ϕ(a) { ∈ } of An. This subset, denoted by one of

ϕA, ϕ(A), is more precisely a 0-definable relation of A. If B A, and ϕ is a formula of S (B), then ϕA is a B-definable relation of⊆ A.

.. Terms

The terms of S are defined recursively as follows.

. Every variable is a term: we use variables xi, where i ω. These variables are more precisely called individual variables,∈ because they stand for individual elements of sets.

  Model theory

. For all n in ω, if F is an n-ary operation-symbol of S , and t0,..., tn−1 are terms of S , then the string

Ft t 0 ··· n−1 is a term of S . In particular, if n = 0, so that F is a constant, then, standing by itself, it is a term. This recursive definition allows theorems about all terms to be proved by induction. A trivial example is that every term that is not a variable begins with an operation-symbol (possibly a constant); slightly less trivial is that every term ends with a variable or a constant. We also want to be able to define functions recursively on the set of terms (of a given signature). This requires knowing that every term is uniquely readable: it can be constructed in only one way. For example, we want to have that, if A Mod(S ), then the terms t of S that use only ∈ variables from the set xi : i < n can be understood as n-ary operations tA on A as follows: { }

A xi (a)= ai, Ft t A(a)= F A(t A(a),...,t A(a)). 0 ··· n−1 0 n−1

This is a valid definition, only if we know that Ft0 ...tn−1 cannot also be analyzed as Fu0 ...un−1 for some terms ui, where in at least one case ui is not ti. But this is true: Ft0 ...tn−1 cannot be otherwise analyzed. Perhaps the simplest way to prove it is by means of a lemma, proved by induction: every term neither is a proper initial segment of another term, nor has a proper initial segment that is a term.

.. Formulas

The atomic formulas of S are of two kinds: equations, which take the form t = u, where t and u are terms of S ; and expressions of the form

Rt t , 0 ··· n−1

 . Definable relations

where n ω, R is an n-ary predicate in S , and the ti are terms of S . Then the∈formulas of S in general are defined recursively: . Atomic formulas are formulas. . If ϕ and ψ are formulas, then so are ϕ and (ϕ ψ). . If ϕ is a formula and x is a variable,¬ then x ϕ is∧ a formula. ∃ Because this definition (like that of terms) is recursive, induction can be used to prove that certain sets of formulas of S contain all formulas of S . For example, induction will be used to establish the lemma called the Tarski–Vaught Test on page  below. However, that lemma also involves truth, which can be understood formally as a function defined recursively on the set of all formulas of S . (See the next subsection.) As with terms, for such a definition to be valid, unique readability of formulas must be established: every formula is built up from its subformulas, but this can happen in only one way. An occurrence of a variable x in a formula is free, unless this occurrence is in a subformula of the form x ϕ: then the occurrrence is bound. This definition would be possible without∃ unique readability, but it would not then be particularly useful. There are some standard abbreviations:

(ϕ ψ) means ( ϕ ψ); ∨ ¬ ¬ ∧¬ (ϕ ψ) means ( ϕ ψ); → ¬ ∨ (ϕ ψ) means ((ϕ ψ) (ψ ϕ)); ↔ → ∧ → x ϕ means x ϕ. ∀ ¬∃ ¬ It is possible to remove some parentheses from formulas without ambigu- ity, if it is understood for example that and take precedence over and , and of two instances of , the one∧ on the∨ right takes precedence.→ Also↔ outer parentheses can be removed.→

.. Truth

Now sentences are defined as before (p. ), as formulas with no free vari- ables; and we can now define truth of sentences in structures. Assuming

  Model theory

t, u, and the ti are terms with no variables, and σ and τ are sentences, but ϕ is formula with the unique free variable x, we have

A  t = u tA = uA, ⇐⇒ A  Rt t (t A,...,t A) RA, 0 ··· n−1 ⇐⇒ 0 n−1 ∈ A  σ A 2 σ, ¬ ⇐⇒ A  σ τ A  σ & A  τ, ∧ ⇐⇒ A  x ϕ(x) for some b in A, A  ϕ(b). ∃ ⇐⇒ Note here that the expressions & and are not parts of formulas; they are abbreviations of the English and⇐⇒and if and only if. We can understand our definition as follows. Given the structure A in Mod(S ), we have now recursively defined, on the set of formulas of S (A), the function ϕ ϕA, where, if ϕ is n-ary, we have 7→ ϕA = a An : A  ϕ(a) . { ∈ } It is worthwhile to note that as ϕ varies here, ϕA ranges over a subset of An with some natural closure properties; and as n varies as well, there are still some good closure properties. Section . is an investigation of these. Before that (as well as later), it is useful to have the notion of substructure, developed in the next section.

. Substructures

B Mod(S ), and A B, and A is closed under the operations of B, then∈ A is the universe⊆ of a substructure of B. This substructure is A, where

RA = An RB, ∩ F A = F B ↾ An for all n-ary predicates R and operation-symbols F of S , for all n in ω. In this case, we write A B. ⊆

 . Substructures

It follows that, if A and B are two structures with the same signature S , and A B, then A B if and only if, for all n in ω, for all n-ary quantifier-free⊆ formulas ϕ⊆of S , for all a in An,

A  ϕ(a) B  ϕ(a). ( ) ⇐⇒ ‡ Suppose now there is just a function h from A to B. Then h is an embedding of A in B if h is an isomorphism from A to a substructure of B. In any case, the diagram of A, denoted by

diag(A), is the set of quantifier-free sentences of S (A) that are true in A. When we consider A as having signature S (A), we may write the structure as

AA.

This is an expansion of A to S (A); and A is then the reduct of AA to S . Now Bh[A] can also be understood as having signature S (A). In this case, the map h above is an embedding if and only if

Bh[A]  diag(A). Thus the structures in which A embeds are precisely (the reducts to S of) the models of diag(A). A theory T of S is axiomatized by a subset Γ of Sn(S ) if

T = Th(Mod(Γ)), equivalently, every model of Γ is a model of T . A universal formula is a formula of the form x ϕ, where ϕ is quantifier- free. If T is a theory, then we denote by ∀

T∀ the theory axiomatized by the universal sentences in T .

Lemma. T∀ is included in the theory of substructures of models of T , that is, A B & B  T = A  T . ⊆ ⇒ ∀

  Model theory

Proof. Suppose ϕ is quantifier-free, and x ϕ is in T . If a An, then a Bn, so B  ϕ(a) and therefore, by ( ),∀A  ϕ(a). Thus A ∈ x ϕ. ∈ ‡ ∀ The converse is given in Theorem  on page  below. If ( ) holds for all formulas ϕ of S , then A is called an elementary substructure‡ of B, and we write A 4 B. A map h from A to B is an elementary embedding of A in B if h is an isomorphism from A to an elementary substructure of B. Thus the structures in which A embeds elementarily are precisely (the reducts to S of) the models of Th(AA). A theory T of a signature S is called model-complete if for all models A of T , the theory of S (A) axiomatized by T diag(A) is complete. ∪ Theorem . A theory is model-complete if and only if, for all of its models A and B, A B = A 4 B. ⊆ ⇒ Proof. Each condition is equivalent to the condition that, for all models A of T , T diag(A) axiomatizes Th(A ). ∪ A The theorem below is a generalization of the theorem published by Löwen- heim in  [] and improved by Skolem in  []: a sentence with a model has a countable model. Skolem’s argument uses what we shall call the Skolem normal form of the given sentence; we shall discuss this in §.. Meanwhile, an example is x y Rxy. If this has a model A, then there is a singulary operation x∀ ∃x∗ on A such that 7→ A  x Rxx∗. ∀ If b A, we can define (b : k ω) recursively by ∈ k ∈ ∗ b0 = b, bk+1 = bk .

Then bk : k ω is countable and is the universe of a substructure of A in which{ ∈x y} Rxy is true. Our own proof of the general theorem will follow the∀ lines∃ of Skolem’s idea; the following lemma will allow us to work in the general situation.

 . Substructures

Lemma (Tarski–Vaught Test). Suppose A B, both having signature S . Then A 4 B, provided that, for all singulary⊆ formulas ϕ of S (A),

B  x ϕ(x) = for some c in A, B  ϕ(c), ∃ ⇒ that is, ϕB = ∅ = ϕB A = ∅. 6 ⇒ ∩ 6 Proof. Under the given condition, we show by induction that for all for- mulas ϕ of S , if ϕ is n-ary and a An, then ∈ A  ϕ(a) B  ϕ(a). ⇐⇒ This is given to be the case when ϕ is atomic (or more generally quantifier- free), and it is easily preserved under negation and conjunction. Suppose it holds when ϕ is an (m + 1)-ary formula ψ. By hypothesis, for all a in An, the following are equivalent:

B  y ϕ(a,y), ∃ for some b in B, B  ϕ(a,b), for some b in A, B  ϕ(a,b), for some b in A, A  ϕ(a,b), A  y ϕ(a,y). ∃ This completes the induction. Theorem  (Downward Löwenheim–Skolem–Tarski). If B Mod(S ), max( S , ω) 6 B , and X B, there is a structure A such∈ that | | | | ⊆ A 4 B, X A, A 6 κ, ⊆ | | where κ = max( X , S , ω). | | | | Proof. There is a subset X′ of B of cardinality no greater than κ such that, for every singulary formula ϕ of S (X), if ϕB is non-empty, then it has an element in X′. By considering formulas a = x, where a X, we see X X′. Now we can form X′′, and so forth; and we can let∈ ⊆ A = X X′ X′′ ... ∪ ∪ ∪

  Model theory

By considering formulas F x = y, we see that A is the universe of a substructure A of B. It is of the required cardinality, and by the Tarski– Vaught Test, it is an elementary substructure of B.

  Ultraproducts and Łoś’s Theorem

. Ideals and filters

For every set Ω, the power-set P(Ω) is a ring in which

sums are symmetric differences: + is △; • products are intersections: is ; • · ∩ the additive identity is the empty set: 0 is ∅; • every element is its own additive inverse: X is just X; • the multiplicative identity is the whole set:−1 is Ω. • It is easy to check this. We note first:

X △ Y = Y △ X, X △ (Y △ Z)=(X △ Y ) △ Z,

(see Figure .) and

Figure .: Symmetric differences of two sets and three sets

X △ ∅ = X, X △ X = ∅.

So (P(Ω), △, ∅) is an abelian group in which every element is its own inverse. (That is, it is an abelian group of exponent 2. It is a standard

  Ultraproducts and Łoś’s Theorem exercise to show that every group of exponent 2 is abelian.) As for multiplication, we have X (Y Z)=(X Y ) Z, ∩ ∩ ∩ ∩ X Y = Y X, ∩ ∩ X Ω= X, ∩ X (Y △ Z)= X Y △ X Z, ∩ ∩ ∩ where multiplication takes notational precedence over addition △ in ∩ the line; see Figure .. So (P(Ω), △, , ∅, Ω) is a ring (commutative ∩

Figure .: Distribution in Boolean rings and unital, as always in these notes). We have in this ring also X X = X. ∩ That is, in the more usual notation of ring-theory, P(Ω) is a ring in which x2 = x. ( ) ∗ Therefore P(Ω) is called a Boolean ring. We shall see in Chapter  that all of the algebraic properties of P(Ω) and its sub-rings follow from their being Boolean rings: the class of Boolean rings is precisely the class of rings that embed in rings of the form P(Ω). This ring P(Ω) has ideals in the usual sense: a subset I of P(Ω) in ideal if and only if ∅ I, ∈ Y I = X Y I, ( ) ∈ ⇒ ∩ ∈ † X I & Y I = X △ Y I. ( ) ∈ ∈ ⇒ ∈ ‡

 . Ideals and filters

However, since X Y Y, ∩ ⊆ X Y X Y = X, ⊆ ⇐⇒ ∩ the condition ( ) is equivalent to † X Y & Y I = X I. ⊆ ∈ ⇒ ∈ Since also X Y = X △ Y △ X Y, ∪ ∩ X △ Y X Y, ⊆ ∪ we can replace the condition ( ) with ‡ X I & Y I = X Y I. ∈ ∈ ⇒ ∪ ∈ That is, a subset I of P(Ω) is an ideal if and only if

∅ I, ∈ X Y & Y I = X I, ⊆ ∈ ⇒ ∈ X I & Y I = X Y I. ∈ ∈ ⇒ ∪ ∈ Thus ideals of P(Ω) are just those subsets that are downwardly closed, • closed under finite unions. • Indeed, taken strictly, the last condition implies non-emptiness, since ∅ is the finite union ∅. See Figure .. Our observations in §. generalize to show that if (Ki : i Ω) is an indexed family of fields, and J is an ideal S ∈ P of i∈Ω Ki, then supp[J] is an ideal of (Ω). Moreover, every ideal I of P(Ω) is supp[J], where Q J = x K : supp(x) I ; { ∈ i ∈ } iY∈Ω and this is an ideal of i∈Ω Ki, in fact the only ideal whose image under J supp[J] is I. So this map establishes a one-to-one correspondence 7→ Q P between the (set of) ideals of i∈Ω Ki and the (set of) ideals of (Ω). There are two standard examplesQ of ideals of P(Ω).

  Ultraproducts and Łoś’s Theorem

Ω b

X Y ∪b

X b b Y

b ∅

Figure .: An ideal of a Boolean ring

. If A Ω, then P(A) is the principal ideal of P(Ω) generated by ⊆ A. It corresponds to the ideal i∈A Ki of i∈Ω Ki. . The set Pω(Ω) of finite subsets of Ω is an ideal of P(Ω), called P Q Q the Fréchet ideal of (Ω), corresponding to the ideal i∈Ω Ki of Ki. i∈Ω P There isQ a notion that is ‘dual’ to the notion of an ideal: A subset F of P(Ω) is a filter if Xc : X F is an ideal, that is, { ∈ } Ω F, ∈ X F & X Y = Y F, ∈ ⊆ ⇒ ∈ X F & Y F = X Y F. ∈ ∈ ⇒ ∩ ∈ See Figure .. Intuitively, elements of an ideal are ‘small’ subsets of Ω; elements of a filter are ‘large’ subsets of Ω.

Since filters easily correspond to ideals, we need not introduce the concept of a filter. Alternatively, once we have filters, we can forget about ideals. I prefer not to forget about ideals, since they are familiar from ring-theory. And yet sometimes it will be useful to think in terms of filters as well.

 . Ideals and filters

Ω b

b b X Y

b X Y ∩

b ∅

Figure .: A filter of a Boolean ring

A filter of P(Ω) is also called a filter on Ω itself. Suppose P is a prime ideal of P(Ω), that is, P is an ideal such that

X Y P & X / P = Y P. ∩ ∈ ∈ ⇒ ∈ The quotient P(Ω)/P is then an integral domain. But it is an integral domain in which ( ) holds, that is, for all elements x, ∗ x2 = x, 0= x2 x = x (x 1). − · − In every integral domain, these equation is solved by 0 and 1, but by no other elements. So P(Ω)/P has only two elements, namely P and Ω+P . A two-element integral domain (even a two-element ring) is a field. So P(Ω)/P is a field, and therefore P is a maximal ideal. Thus all prime ideals of P(Ω) are maximal. Since in P(Ω) we have

X △ Ω= Xc, we have that X and Xc are always in different cosets of P , that is,

X P Xc / P. ∈ ⇐⇒ ∈

  Ultraproducts and Łoś’s Theorem

The filter Xc : X P that is dual to P is called an ultrafilter. As a filter, it is the{ set of∈ complements} (in Ω) of elements of P ; as an ultrafilter, it is the complement (in P(Ω)) of P .

. Reduced products

Suppose (Ai : i Ω) is an indexed family of structures with a common S ∈ signature . An element of the Cartesian product i∈Ω Ai is a tuple (ai : i Ω), where ai Ai; we may write this tuple simply as a. Let the ∈ ∈ Q product i∈Ω Ai be called also B. This is the universe of a structure Q Ai iY∈Ω or B of S , where for all n in ω, for all n-ary operation-symbols F and predicates R of S ,

F B(a)=(F Ai (a ): i Ω), RB = RAi . i ∈ iY∈Ω Here notation is as in §, so that a Bn, which means a =(a0,...,an−1), where ak = (ak : i Ω); and a =∈ (a0,...,an−1). See Figure .. Note i ∈ i i i

a a0 a1 a2 ... q qqq 0 0 0 0 a = a0 a1 a2 ...... n−1 n−1 n−1 n−1 a = a0 a1 a2 ...

Figure .: Notation for sequences of tuples that, if RAi = ∅ for even one index i, then RB = ∅. Now let I be an ideal of P(Ω). We shall define a quotient

Ai/I iY∈Ω

 . Reduced products or B/I. First, we define two elements of B to be congruent modulo I if they disagree only on a small set of indices, that is, a set belonging to I: i Ω: a = b I a b (mod I). { ∈ i 6 i}∈ ⇐⇒ ≡ Equivalently, if F is the dual filter of I (that is, F = Xc : X I , then two elements of B are congruent if and only if they agree{ on a∈ large} set of indices, that is, a set belonging to F : i Ω: a = b F a b (mod I). { ∈ i i}∈ ⇐⇒ ≡ We may write: a/I for the congruence-class b: a b (mod I) , • B/I for the set of these congruence-classes,{ ≡ } • a/I for (a0/I,...,an−1/I). • (Or one could write F for I here.) We want to define a structure B/I of S with universe B/I. Then for all n-ary operation-symbols F of S , we should define F B/I (a/I)= F B(a)/I. However, we must check that this is a valid definition. We shall do this presently; meanwhile, for an n-ary predicate R of S , what should RB/I be? By one generalization of the definition in §. of the ordering of ω R /I, we should define RB/I as a/I : a RB . { ∈ } However, by what we noted above, this relation is empty if RAi = ∅ for even one index i. The empty relation is still a relation, but it is not what we want here. The ‘correct’ definition of RB/I is given as follows.

Theorem . For all indexed families (Ai : i Ω) of structures with common signature S , there is a structure B/I in∈Mod(S ), with universe ω i∈Ω Ai/I, such that all n in , for all n-ary operation-symbols F and predicates R of S , Q F B/I (a/I)= F B(a)/I, RB/I = a/I : i: a / RAi I . { i ∈ }∈ If RAi = ∅ for each i in Ω, then  6 RB/I = a/I : a RB . { ∈ }

  Ultraproducts and Łoś’s Theorem

Proof. We need only check that the definition of F B/I is valid. Suppose

a b (mod I), ≡ that is, a/I = b/I, that is, ak/I = bk/I for each k in n. This means i: ak = bk I for each k in n. But then { i 6 i }∈ i: ak = bk I. { i 6 i }∈ k[∈n We have also

i: F Ai (a ) = F Ai (b i: ak = bk , { i 6 i}⊆ { i 6 i } k[∈n so i: F Ai (a ) = F Ai (b ) I and therefore { i 6 i }∈ F B(a) F B(b) (mod I). ≡

The structure B/I is a reduced product of the family (A : i Ω). i ∈ We can understand each element a of i∈Ω Ai as a new constant, to be interpreted in each Ai as ai, and in B as a/I. Then the definition Q RB/I = a/I : i: a / RAi I { i ∈ }∈ in the theorem just means 

B/I  Ra i: A  Ra I, ⇐⇒ { i ¬ }∈ and the definition F B/I (a/I)= F B(a)/I has the meaning of

B/I  F a = b i: A  (F a = b) I. ⇐⇒ { i ¬ }∈ The formulas Rx and F x = y here are unnested atomic formulas. (So is x = y.) Our equivalences have the form

B/I  σ i: A  σ I, ( ) ⇐⇒ { i ¬ }∈ §

 . Reduced products which can be written also as

B/I  σ i: A  σ F. ⇐⇒ { i }∈ So this is true by definition when σ is an unnested atomic sentence (in the expanded signature S (B)). For which other σ is it true? Lemma. The equivalence ( ), that is, § B/I  σ i: A  σ I, ⇐⇒ { i ¬ }∈ considered as a function of σ, is preserved under conjunction: if it holds when σ is τ or ρ, then it holds when σ is τ ρ. ∧ Proof. From the definition of an ideal, we have

X Y & Y I = X I, ⊆ ∈ ⇒ ∈ X I & Y I = X Y I. ∈ ∈ ⇒ ∪ ∈ Together these imply the converse of the latter; so we have

X I & Y I X Y I. ∈ ∈ ⇐⇒ ∪ ∈ If now ( ) holds when σ is τ or ρ, then the following are equivalent: § B/I  τ ρ, ∧ B/I  τ & M  ρ, i: A  τ I & i: A  ρ I, { i ¬ }∈ { i ¬ }∈ i: A  τ i: A  ρ I, { i ¬ } ∪ { i ¬ }∈ i: A  τ ρ I, { i ¬ ∨¬ }∈ i: A  (τ ρ) I. { i ¬ ∧ }∈ Lemma. The equivalence ( ), that is, § B/I  σ i: A  σ I, ⇐⇒ { i ¬ }∈ considered as a function of σ, is preserved under quantification: if it holds when σ is ψ(a), for all parameters a, for some singulary for- mula ψ (possibly with parameters),

  Ultraproducts and Łoś’s Theorem then it holds when σ is xψ(x). ∃ Proof. Under the given hypothesis, the following are equivalent: B/I  xψ(x), ∃ B/I  ψ(a) for some a in B, i: A  ψ(a ) I for some a in B. { i ¬ i }∈ For all a in B, we have i: A  ψ(a ) i: A  xψ(x) . { i ¬ i } ⊇ { i ¬∃ } Moreover, for some choice of a, this inclusion is an equality. This yields the result.

. Ultraproducts

Suppose P is a prime ideal (hence a maximal ideal) of P(Ω). Then the reduced product B/P is called more precisely an ultraproduct of (A : i Ω). There is a trivial example: For some j in Ω, let P be the i ∈ principal ideal (Ωr j ) of P(Ω), that is, P = X Ω: j / Ω . Then { } { ⊆ ∈ } Ai/P ∼= Aj. iY∈Ω All ultraproducts are thus if Ω is finite; but if Ω is infinite, then the Fréchet ideal of P(Ω) (that is, the ideal consisting of all finite subsets of Ω) is a proper ideal, so it may be included in P , which is therefore not principal. Immediately we have: Lemma. If P is a prime ideal of P(Ω), then the equivalence B/P  σ i: A  σ P, ⇐⇒ { i ¬ }∈ as a function of σ, is preserved under negation.

So we are assuming the Prime Ideal Theorem, that every proper ideal of a ring is included in a prime ideal. This is apparently weaker than the Maximal Ideal Theorem, that every proper ideal of a ring is included in a maximal ideal; this is equivalent to the Axiom of Choice. Some discussion and references are found in [, §., pp. –].

 . Cardinality

The three lemmas of this chapter together yield: Theorem  (Łoś). If P is a maximal ideal of P(Ω), then B/P  σ i: A  σ P ⇐⇒ { i ¬ }∈ holds for all σ (with parameters).

As a special case, if each Ai is the same structure A, so that in particular Ω i∈Ω Ai is the Cartesian power A , then the ultraproduct B/I (that is AΩ/P ) is an ultrapower of A. The diagonal embedding a (a: i Ω) Qof A in B/I is now an elementary embedding, that is, for all7→ sentences∈ σ with parameters from A (or more precisely from the image of A in AΩ), B/I  σ A  σ. ⇐⇒ Considering the embedding as an inclusion (that is, identifying A with its image in B/I), we may write then A 4 B/I.

. Cardinality

By the theorem below, a non-principal ultrapower C of a countably infi- nite structure A is uncountable. By the Downward Löwenheim–Skolem– Tarski Theorem, in a countable signature, there will then be a countable structure B such that A B C. ≺ ≺ Indeed, B may be chosen to include A x for some x in CrA. However, even though A is then a proper substructure∪{ } of B, these two may be isomorphic. However, this is not the case when A is (N, +, ). Thus countable non-standard models of arithmetic exist. A more illuminating· construction of such models is given in §. below. The following is a special case of [, Thm ..(a)] (and is said to be found in Frayne, Morel, and Scott []). The usual reference is [] although the theorem is not given clearly there. I have a printout of this article, but have not sorted through all of its many basic results to find this one. It should be noted that the article has a ‘correction’ [], which merely refines the account of Tarski’s contribution to the subject (as well as taking some of the credit away from Frayne).

  Ultraproducts and Łoś’s Theorem

Theorem . For all signatures S , for all A in Mod(S ), for all singulary formulas ϕ of S (A), for all non-principal prime ideals P of P(ω),

ω ω ω 6 ϕ(A) = ϕ(A /P ) = ϕ(A) . | | ⇒ | | | | In particular, if A is countable, then all infinite definable relations of Aω/P have the cardinality of the continuum.

Proof. For all a in Aω, if a/P ϕ(Aω/P ), then by Łoś’s Theorem ∈ i ω: a / ϕ(A) P. { ∈ i ∈ }∈

Then we may assume this set i ω: ai / ϕ(A) is actually empty. More precisely, there is a′ in ϕ(A)ω{such∈ that a/P∈ = a}′/P . More precisely still, there is an injection a/P a′ from ϕ(Aω/P ) to ϕ(A)ω. This shows 7→ ω ω ϕ(A /P ) 6 ϕ(A) . | | | | For the reverse inequality, it is enough to find a function x x∗ from ϕ(Aω) to itself such that 7→

x = y = x∗/P = y∗/P, 6 ⇒ 6 that is, x = y = i: x∗ = y∗ / P. 6 ⇒ { i 6 i } ∈ Now x = y means x = y for some i in ω. If 6 i 6 i j : x∗ = y∗ ω r i, ( ) { j 6 j }⊇ ¶ ∗ ∗ ∗ ∗ that is, if j : xj = yj i, then since i P , we have j : xj = yj / P . So it is enough{ that }⊆ ∈ { 6 } ∈

x = y & i 6 j = x∗ = y∗. ( ) i 6 i ⇒ j 6 j k ∗ We can achieve this by letting xj be an injective function of (x0,...,xj ), for each j in ω. We can do this, if ϕ(A) is infinite. Indeed, for each i in i+1 ω, let µi be an injection from ϕ(A) to ϕ(A). Now we can define

∗ xi = µi(x0,...,xi).

 . Cardinality

Let us try to generalize this argument, replacing ω with an arbitrary infinite index-set Ω. In the condition ( ), the element ω r i of the dual ¶ filter F of P will be replaced by some element Xi of the dual filter. Then ( ) becomes k x = x & j X = x∗ = y∗. i 6 j ∈ i ⇒ j 6 j ∗ So xj should be an injective function of (xi : j Xi), for each j in Ω. For this, it is enough if the sets ∈

i Ω: j X { ∈ ∈ i} are finite. An ultrafilter F on Ω (that is, an ultrafilter of P(Ω) is called regular if it has such elements Xi for all i in Ω.

It is easy to show that there are regular ultrafilters on Pω(Ω) (that is, regular ultrafilters of P(Pω(Ω)). For, if i Pω(Ω), we need only define ∈ X = j Pω(Ω): i j . i { ∈ ⊆ } Since Xi Xj = Xi∪j, the Xi do generate a filter on Pω(Ω). The filter is proper,∩ since i X , so none of the X is empty. Moreover, ∈ i i i Pω(Ω): j X = i Pω(Ω): i j = P(j), { ∈ ∈ i} { ∈ ⊆ } which is finite. So there are regular proper filters, and hence regular ultrafilters, on Pω(Ω). Since Pω(Ω) has the same cardinality as Ω (assuming this is infinite), there are regular ultrafilters on Ω.

We shall use an index-set of the form Pω(Ω) in proving the Compactness Theorem (Theorem ) on page  below.

  Simple applications

. Arrow’s Theorem

This section is inspired by Sasha Borovik’s article []. We consider an index-set Ω as a set of voters. Each voter i in Ω is called on to assign a linear ordering

D = ∅, 6 X D = Xc / D. ∈ ⇒ ∈ We also require that additional votes for a particular ordering can only help that ordering: X D & X Y Ω = Y D. ∈ ⊆ ⊆ ⇒ ∈ Hence in particular Ω D. We require voting to be decisive: ∈ X / D = Xc D. ∈ ⇒ ∈ If A consists of just two candidates, this is all we need. Then D is not necessarily an ultrafilter on Ω; for it need not be closed under intersec- tions. Indeed, in the ‘democratic’ case, if Ω has a finite number 2n 1 −

 . Compactness

c

c

Figure .: An election with three candidates of members, then D will be X P(Ω): X > n ; this is definitely not closed under intersections unless{ ∈n = 1. | | } But now suppose A contains three distinct candidates, a, b, and c; and let i: a< b = A, i: b< c = B. { i } { i } Suppose both A and B are in D. Then we must conclude a

i: c< a< b = A r B, { i i } i: b< c< a = B r A, { i i } i: c< b< a =(A B)c. { i i } ∪ See Figure .. Thus we must have A B D. Therefore D is an ultrafilter on Ω. If Ω is finite, then D must∩ be a∈ principal ultrafilter: that is, one voter decides everything, and the system is a dictatorship.

. Compactness

Theorem  (Compactness). Suppose Γ Sn(S ), and every finite subset of Γ has a model. Then Γ itself has a model.⊆

  Simple applications

Proof. Assuming A∆  ∆ for each ∆ in Pω(Γ), we shall find an ultrafilter F on Pω(Γ) such that

A∆/F  Γ. ( ) P ∗ ∆∈Yω(Γ) This just means, by Łoś’s Theorem, that for each σ in Γ,

∆ Pω(Γ): A  σ F. { ∈ ∆ }∈ But we have

∆ Pω(Γ): σ ∆ ∆ Pω(Γ): A  σ . { ∈ ∈ } ⊆ { ∈ ∆ } Let the former set be denoted by [σ]; more generally, if ∆ Pω(Γ), let ∈ [∆] = Θ Pω(Γ): ∆ Θ . { ∈ ⊆ } Then

∆ [∆], [∆] [Θ] = [∆ Θ]. ∈ ∩ ∪ Consequently the sets [∆] generate a filter on P(Pω(Γ)). Let it be included in the ultrafilter F , and let B be the ultraproduct of (A : ∆ ∆ ∈ Pω(Γ)) with respect to F . For every σ in Γ, we have [σ] F , and ∈ for every ∆ in [σ], we have A∆  σ. Then by Łoś’s Theorem, we have ( ). ∗ Now we can establish a complement to Theorem  (p. ): Theorem  (Upward Löwenheim–Skolem–Tarski). If A is an infinite structure with signature S , and max( A , σ ) 6 κ, then there is a struc- ture B such that | | | |

A 4 B, B = κ. | |

Proof. Let C be a set cα : α<κ be a set of new constants, all distinct. By Compactness, the set{ }

Th(A ) c = c : α<β<κ A ∪ { α 6 β }

 . Compactness

of sentences has a model DA∪C . By construction, this model has car- dinality at least κ. By the downward version of the theorem, D has an elementary substructure B of size κ such that A B. Since also A 4 D, the structure B is as desired. ⊆

This theorem yields an easy test for completeness of theories. For an infinite cardinal κ, a theory is κ-categorical if all of its models of size κ are isomorphic to one another.

Theorem  (Łoś–Vaught Test). If a theory T of a signature S has models, but no finite models; S 6 κ; and T is κ-categorical; then T is complete. | |

Proof. If T contains neither σ nor σ, then both T σ and T σ have models, which must be infinite.¬ Then by the Löwenheim–Sk∪ {¬ } olem–∪ { } Tarski theorems (both upward and downward forms may be needed), each of the two sets has a model of cardinality κ; but these two models cannot be isomorphic to one another.

Theorem . The theory of algebraically closed fields of characteristic 0 is com- • plete. For all primes p, the theory of algebraically closed fields of charac- • teristic p is complete.

Proof. None of these theories has no finite models. Every algebraically closed field is determined up to isomorphism by its characteristic and its transcendence-degree. If κ is uncountable, then a field with transcendence- degree κ has cardinality κ. Now the Łoś–Vaught Test applies.

Similarly we have the following (see page  above):

Theorem . The theory of algebraically closed fields is model-complete.

Proof. If T is this theory, K  T , and K < κ, then T diag(K) is κ-categorical, but has no finite models. | | ∪

  Simple applications

We can also now prove the converse of the lemma on page  above.

Theorem . For all theories T , the models of T∀ are precisely the substructures of models of T .

Proof. Assuming A  T∀, we want to show T diag(A) has a model. By Compactness, and since diag(A) is closed under∪ conjunction, it is enough to show T ϑ(a) has a model whenever ϑ is a quantifier-free formula of S and A∪ { ϑ(a}). If it has no model, then T ϑ(a), so (since no entry of a is in S ) T x ϑ(x), and therefore A⊢ ¬ x ϑ(x), which is absurd. ⊢∀ ¬ ∀ ¬

In particular, when T is just field-theory, then T∀ is the theory of integral domains: see page  below.

. Elementary classes

In [] Łoś defined ultraproducts (but not by that name) in order to state the following algebraic test for being an elementary class of structures.

Theorem . A subclass of Mod(S ) is elementary if and only if it contains: every structure that is elementarily equivalent to a member, and • every ultraproduct of members. •

Proof. The ‘only if’ direction is the easier. An elementary class is the class of models of some theory T . If the class is , and A , and A B, then B  T , so B . If A : i Ω ,K then A  ∈T Kin each ≡ ∈K { i ∈ }⊆K i case, so every ultraproduct of the Ai is a model of T , by Łoś’s Theorem. The more difficult direction is ‘if’. Suppose is a non-elementary sub- class of Mod(S ). Then there is a model B of KTh( ) that does not belong to . However, every element σ of Th(B) has a modelK in , since other- wiseK σ would be in Th( ). Therefore every finite subset ∆Kof Th(B) has ¬ K a model A∆ in (since otherwise the negation of the conjunction of the members of ∆ wouldK be in Th( )). By (the proof of) the Compactness K

 . A countable non-standard model of arithmetic

Theorem, some ultraproduct of (A∆ : ∆ Pω(Th(B))) is elementarily equivalent to B. ∈

. A countable non-standard model of arithmetic

By arithmetic we mean the theory of (ω, +, ) or of (ω, +, , 0, 1, 6); it makes little difference, since · · ) 6 is definable in (ω, +, ) by the formula z x + z = y, ) 0 is definable by y y·+ x = y, ∃ ) {1} is definable by 0∀

(ω, +, )  y ϕ(f(i),y) (ω, +, )  ϕ(f(i),g(i)). · ∃ ⇐⇒ · Indeed, g can be such that g(i) is the least b such that (ω, +, )  ϕ(f(i),b), if such b exist; and otherwise g(i) = 0. Then g is defined·

  Simple applications by the formula

(ϕ(f(x),y) ( z (ϕ(f(x),z) y 6 z))) ( z ϕ(f(x),z) y = 0). ∧ ∀ → ∨ ∀ ¬ ∧ It follows by the Tarski–Vaught Test (page ) that

A 4 B; therefore, since (ω, +, ) A, we have · ⊆ (ω, +, ) A. · ≺ Indeed, we now have that the following are equivalent:

B  y ϕ(f,y), ∃ i: (ω, +, )  y ϕ(f(i),y) F, { · ∃ }∈ i: (ω, +, )  ϕ(f(i),g(i)) F, { · }∈ B  ϕ(f,g).

Now the Tarski–Vaught Test applies. This construction of A is apparently due to Skolem.

I take it from Bell and Slomson [, Ch. , §].

  Gödel’s Completeness Theorem

A sentence that is true in all structures of its signature can be called valid sentence or a validity. It is easy in principle to show that a sentence is not valid: just exhibit a model of its negation. But if a sentence is valid, how can this be shown? We cannot simply verify the sentence in all structures of its signature, since there will be infinitely many of these structures. The method of formal proof is a finitary alternative. We shall show by means of ultraproducts that this method always works in principle. This result is Gödel’s Incompleteness Theorem.

. Formal proofs

A formal proof is just a (finite) list of sentences such that each sentence on the list is either

) an axiom, or ) derivable from sentences earlier in the list by means of a rule of inference.

We choose the axioms and rules of inference to serve our needs; taken all together, they constitute a proof-system. In his doctoral dissertation of , Gödel [] gave a particular proof-system, obtained from the Principia Mathematica [] of Russell and Whitehead. The first four of Gödel’s axioms, or rather schemes of axioms, are found on page , Chapter , of the Principia Mathematica. Recall that, by our convention

As Gödel notes, there was a fifth axiom, ϕ ∨ (ψ ∨ χ) → ψ ∨ (ϕ ∨ χ), but apparently Bernays showed it to be redundant. For us, each of the four formulas given here represents infinitely many axioms, since ϕ, ψ, and χ can be any formulas. It should be noted that Russell and Whitehead were involved in creating formal logic; our way of understanding formulas was not yet fully developed. For an amusing fictionalized account of Russell’s interactions with Gödel, see [].

  Gödel’s Completeness Theorem on symbolic precedence given on page , takes precedence over , and of two instances of , the one on the right∨ takes precedence. → → ) ϕ ϕ ϕ, ∨ → ) ϕ ϕ ψ, → ∨ ) ϕ ψ ψ ϕ, ∨ → ∨ ) (ϕ ψ) χ ϕ χ ψ. → → ∨ → ∨ The primitive Boolean connectives here are actually and ; so ϕ ψ should be understood as an abbreviation of ϕ ψ. In∨ Chapter¬  of→ the Principia (at . and ., pp. –) are¬ ∨ found Gödel’s next two axioms: ∗ ∗

) x ϕ ϕx, ∀ → y ) x (ϑ ϕ) ϑ x ϕ. ∀ ∨ → ∨∀ x In the former scheme, ϕy is the result of replacing every free occurrence of x in ϕ with y. In the latter scheme, x must not occur freely in ϑ. The primitive quantifier can be taken as , so that x ψ will be an abbreviation of x ψ. Finally, equality is∀ treated in two∃ axioms, found in Chapter  of¬∀ the¬Principia (at . and ., pp. –): ∗ ∗ ) x = x,

) x = y ϕ ϕx. → → y Here y should not occur in ϕ, or at least there should be no subformula y ψ in which there is an occurrence of x that is free as an occurrence in ∀ϕ.

The rules of inference are three:

See the previous note. For Gödel, there were just six axioms in all, using propo- sitional variables where I have put ϕ, ψ, and χ; and using a functional variable where I have put ϑ. Then in addition to the three rules of inference given below, there was a fourth, allowing propositional and functional variables to be replaced by formulas in our sense. See the previous note on Gödel’s fourth rule of inference (which was actually third on his list). Gödel notes, ‘Although Whitehead and Russell use these rules throughout their derivations, they do not formulate all of them explicitly.’

 . Formal proofs

Detachment: From ϕ and ϕ ψ may be inferred ψ. Generalization: From ϑ may→ be inferred xϑ. Change of variables: ‘Individual variables∀ (free or bound) may be re- placed by others, so long as this does not cause overlapping of the scopes of variables denoted by the same sign’ [, p. ].

Again, a formal proof is a finite list of formulas such that each formula on the list is an axiom or else is derived from previous formulas on the list by means of a rule of inference. The last formula on the list is then said to be provable. If ϕ is provable, we may express this by

ϕ. ⊢ Note that in fact every formula in a formal proof is provable, because every initial segment of a formal proof is still a formal proof.

A generalization of a formula ϕ is a formula x ϕ, where all free vari- ables of ϕ occur in x. Then we can generalize∀ the notion of validity by saying that an arbitrary formula is valid if some (and hence every) generalization of it is true in every structure of its signature.

It should be clear that every provable formula is valid. Gödel proves the converse: this is his Completeness Theorem.

Before Gödel, a completeness theorem for propositional logic was known. Propositional formulas are, strictly, not formulas as defined in §.. above; but they can be understood as formulas in which:

) the place of atomic formulas is taken by propositional variables; ) no quantifier or is used. ∃ ∀ In particular, there are no individual variables in a propositional formula, but only propositional variables. A structure for propositional logic as- signs a truth-value to each of these propositional variables. Then a propo- sitional formula is true or false in the structure, according to the relevant

Detachment is not Gödel’s name for this rule; he (or more precisely his translator) calls it the Inferential Schema. Gödel’s reference for this is Bernays from ; but the theorem can be found in Post’s  article [].

  Gödel’s Completeness Theorem parts of the definition of truth of sentences (on page ), namely:

A  σ A 2 σ, ¬ ⇐⇒ A  σ τ A  σ & A  τ. ∧ ⇐⇒ Strictly, since is now primitive instead of , we should replace the latter rule with ∨ ∧ A  σ τ A  σ or A  τ. ∨ ⇐⇒ We may treat the truth-value true as 1, and false as 0. Then a proposi- tional formula F in an n-tuple (P0,...,Pn−1) of propositional variables determines an n-ary operation Fˆ on 2, where if e 2n, then Fˆ(e) is the truth-value of F in any propositional structure that∈ assigns the value ei to Pi when i < n. This operation Fˆ can be described completely in a truth-table. If the operation is identically 1, then F is a (proposi- tional) tautology. The first four axiom-schemes above, along with the inference-rule of Detachment, constitute a proof-system for propositional logic in which every tautology is provable. This is possibly not an excit- ing theorem, since there is already an algorithm for determining whether a formula is a tautology: just write out its truth-table.

A tautology in general can be understood as resulting from a proposi- tional tautology by replacing each occurrence of a propositional variable with the same formula of some signature S , for all propositional vari- ables occurring in the propositional tautology. Evidently tautologies in this broader sense are valid. They are therefore provable, by the com- pleteness theorem of propositional logic. If we do not want to bother to prove this completeness theorem, we can just introduce, as new axioms, all tautologies, since again there is an algorithm for determining which formulas these are.

If ϕ is a formula, we may define a formal proof from ϕ as a formal proof in the earlier sense, but with ϕ treated as an axiom, and with generalization of free variables in ϕ not allowed. If ψ is provable from ϕ in this sense, we may write ϕ ψ; ⊢ The restriction is to ensure that, if ψ is provable from ϕ, then the formula ϕ → ψ is valid, while the converse is still true.

 . Completeness by ultraproducts to give it a name, we may call this a sequent. For example, by Detach- ment, the axiom x ϕ ϕx gives us the sequent ∀ → y x ϕ ϕx, ( ) ∀ ⊢ y ∗ because the sequence x ϕ, x ϕ ϕx, ϕx. ∀ ∀ → y y is a formal proof from x ϕ. Likewise, the axiom x (ϑ ϕ) ϑ x ϕ gives us ∀ ∀ ∨ → ∨∀ x (ϑ ϕ) ϑ x ϕ. ( ) ∀ ∨ ⊢ ∨∀ † (Recall that x must not be free in ϑ.)

. Completeness by ultraproducts

Suppose σ is an arbitrary sentence. We want to show that either σ is provable, or else its negation has a model, in fact a countable model. We make several simplifying assumptions: . No operation-symbols occur in σ. . The sign = of equality does not occur in σ. . For some positive integers p and q, for some quantifier-free (p + q)- ary formula ϕ, σ is the sentence x y ϕ(x, y), ∃ ∀ where x and y are respectively a p-tuple and a q-tuple of variables. The justification of these assumptions does not involve ultraproducts, so it is relegated to a later section, §..

Let V be a countably infinite set vk : k ω of individual variables. The power V p is countable, so we may{ assume∈ } V p = x : k ω . { k ∈ } The ensuing argument is based mainly on that of Bell and Slomson [, Ch. , §]. These writers cite J.N. Crossley for the suggestion of introducing ultraproducts to Gödel’s original argument. Church [, §] explicates Gödel’s original argument more faithfully.

  Gödel’s Completeness Theorem

Now we may suppose y : k ω V q, { k ∈ }⊆ where if k<ℓ, then yk and yℓ have no entries in common, and if j 6 k, then xj and yk have no entries in common. We now denote

ϕ(xk, yk) by ϕk,

ϕ0 by ϑ0, ϕ ϑ by ϑ , k+1 ∨ k k+1 x y ϑ by σ . ∀ 0 ···∀ k k k

That is, ϑk is defined recursively in k as shown; and σk is a generalization of ϑk. We shall prove that, for all k in ω,

σ σ. k ⊢

To this end, we note first that since no entry of yk+1 appears in ϑk, we have, as a special case of the sequent ( ), † y (ϑ ϕ ) ϑ y ϕ , ∀ k+1 k ∨ k+1 ⊢ k ∨∀ k+1 k+1 that is (by definition of ϑk+1),

y ϑ ϑ y ϕ . ∀ k+1 k+1 ⊢ k ∨∀ k+1 k+1 By the sequent ( ) (used repeatedly), we have ∗ σ y ϑ . k+1 ⊢∀ k+1 k+1 Therefore, by stringing together the (short) proofs of the last two se- quents, we have σ ϑ y ϕ . ( ) k+1 ⊢ k ∨∀ k+1 k+1 ‡ By repeated use of the axiom allowing removal of universal quantifiers, we have x y ϕ(x, y) y ϕ ; ⊢∀ ¬∀ → ¬∀ k+1 k+1 therefore, by contraposition (justified by completeness for propositional logic), y ϕ x y ϕ(x, y), ⊢∀ k+1 k+1 →∃ ∀

 . Completeness by ultraproducts that is, y ϕ σ. ⊢∀ k+1 k+1 → Combining this with ( ) gives ‡ σ ϑ σ k+1 ⊢ k ∨ and therefore (by Generalization mainly)

σ σ σ. k+1 ⊢ k ∨ Thus if σ σ, then σ σ. Since by change of variables we have k ⊢ k+1 ⊢ σ σ k ⊢ when k = 0, by induction we now have this for all k in ω. Now we can show that either σ is provable, or σ has a model. to do so, ¬ we consider two cases. The first one is easy to dispose of: If some σk is provable, then so is σ itself, and we are done.

So now let us suppose that no σk is provable. Then no ϑk is provable; so it must not be a tautology.

Since ϑk is quantifier-free, but not a tautology, there must be a truth- assignment to its atomic subformulas that makes ϑk false. We can extend this to a truth-assignment F to all atomic formulas in variables from V (that is, in the variables vi) with predicates occurring in ϕ. Since none of those predicates is =, the truth-assignment F determines a structure Ak whose universe is ω such that, for each n in ω, for each n-ary predicate R occurring in ϕ,

RAk = (i(0),...,i(n 1)) ωn : F (Rv v ) = 1 . { − ∈ i(0) ··· i(n−1) }

If we now treat each variable vi as a constant whose interpretation in Ak is i, then we have

A  Rv v F (Rv v ) = 1. k i(0) ··· i(n−1) ⇐⇒ i(0) ··· i(n−1) Then by construction A  ϑ . k ¬ k

  Gödel’s Completeness Theorem

Suppose k 6 ℓ. Then ϑ ϑ , k ⊢ ℓ so that also ϑ ϑ , and hence ¬ ℓ ⊢¬ k A  ϑ . ℓ ¬ k Thus for all j in ω, i ω: A  ϑ j. { ∈ i j}⊆ Now let C be a non-principal ultraproduct of the Ai. It follows that, for all j in ω, C  ϑ . ¬ j (Here vi is interpreted as the image of i under the diagonal map.) Hence the conjuncts of the ϑ are true in C: ¬ j C  ϕ(x , y ). ¬ j j Since we have no operation-symbols in our signature, every subset of C is the universe of a substructure of C. Let B be the image of ω under the diagonal map in C. Since ϕ is quantifier-free, and the interpretations of all of the variables are now in B, we have

B  ϕ(x , y ), ¬ j j B  y ϕ(x , y). ∃ ¬ j But we have arranged things so that every element of Bp is the interpre- tation of some xj. Therefore B  x y ϕ(x, y), ∀ ∃ ¬ that is, σ is false in B.

. Completeness by König’s Lemma

Gödel himself does not use an ultraproduct explicitly in his argument, but he can be understood to create the structure B (with universe ω) as follows. Let (α : k ω) be a list of all of the atomic formulas appearing k ∈ I am guided by Church’s version of Gödel’s argument here. See below.

 . Completeness by König’s Lemma

in the formulas ϕℓ. We define B by determining in each case whether αk is to be true in B (again with variable vi understood as i). This determination can be made recursively as follows. First, we let B  α i: A  α = ω 0 ⇐⇒ |{ i 0}| (that is, α0 is true in B if and only if the set of i such that α0 is true in Ai is infinite), and then B  α i: α Ai = α B & A  α = ω, 1 ⇐⇒ |{ 0 0 i 1}| B  α i: α Ai = α B & α Ai = α B & A  α = ω, 2 ⇐⇒ |{ 0 0 1 1 i 2}| Ai B and so on, where αj = αj means simply that αj is alike true or false in Ai and B. Strictly the definition is by strong or well-ordered recursion, requiring only the single condition

B  α i: α Ai = α B & A  α = ω. k ⇐⇒ |{ j j i k}| j

i: α Ai = α B = ω. ( ) |{ j j }| § j

  Gödel’s Completeness Theorem

If both of these sets are infinite, then the question of whether B  αk can be decided arbitrarily. We decided above that αk would be true in B in this case. However, if at the beginning we had chosen an ultrafilter D on ω, then we could just define

B  α i: A  α D. k ⇐⇒ { i k}∈

Gödel himself is not explicit about how he obtains B. His editor van Heijenoort detects an allusion to König’s Lemma. There are more than one theorem called by this name, but probably what is meant is the following [, Lemma II.., p. ]. A tree is a (partially) ordered set such that, for every element a, the subset x: x

i: α Ai = e { j j } j

 . Arbitrary formulas

(0, 0, 0) (0, 0) (0, 0, 1) (0) (0, 1, 0) (0, 1) (0, 1, 1) ∅ (1, 0, 0) (1, 0) (1, 0, 1) (1) (1, 1, 0) (1, 1) (1, 1, 1)

Figure .: A complete binary tree

. Arbitrary formulas

We have to justify the assumptions about σ made at the beginning of §..

.. Operation-symbols

What we call relations, Gödel calls functions; but he has no symbols for what we call operations. However, even if we do use them, we can dispose of them as follows. Suppose, for some n-ary operation-symbol F , there were an atomic subformula α of σ featuring a term Ft t . 0 ··· n−1 Introducing a new (n + 1)-ary predicate RF , we could replace the term Ft t in α with a new variable x, obtaining an atomic formula α′. 0 ··· n−1

  Gödel’s Completeness Theorem

We could then replace α in σ with the formula x (α′ R t t x), ∃ ∧ F 0 ··· n−1 obtaining σ′. Then σ would be logically equivalent to σ′ x y z (Rxy (Rxz y = z)). ∧∀ ∃ ∀ ∧ → We should show that σ is provable from this equivalent sentence.

.. Equality

Suppose no operation-symbol occurs in σ, but the sign = of equality does occur. We have to deal with the requirement that this sign is inter- preted in every structure as equality itself (and not merely an equivalence- relation). We introduce a new binary predicate , and we replace each occurrence of = in σ with this new predicate , obtaining≡ a new sentence ≡ σ. Now let (R0,...,Rm) be a list of all predicates (including ) occurring in σ′, and let σ′′ be the sentence ≡

σ′ x y x y (R x R y ) . ∧∀ ∀ ≡ → j j → j j j6m ^
(Here xj and yj are initial segments, of appropriate length, of x and y respectively; and x and y are long enough to make this possible.) One shows that σ is provable from σ′′. Also, if A  σ′′, then A is an equivalence-relation on A, and the set of equivalence-classes≡ is the universe of a model of σ.

.. Skolem normal form

Every formula ϕ is equivalent to a formula ϕˆ of the same signature in prenex normal form, that is, with all quantifiers in front. For example, if x is not free in ϑ, then x ϕ ϑ x (ϕ ϑ), x ϕ ϑ x (ϕ ϑ). ∀ ∧ ∼∀ ∧ ∀ ∨ ∼∀ ∨ We shall show that, for every formula ϕ, there is a sentence ϕ˜, possibly with new predicates, with the following two properties.

 . Arbitrary formulas

. ϕ˜ has the form x y ϑ, where ϑ is quantifier-free. . If ϕ˜ is valid, then∃ so∀ is ϕ; but if ϕ˜ has a model, then ϕ will be satisfied in that model (that is, it¬ will define a nonempty¬ subset of that model). The sentence ϕ˜ is a Skolem normal form for ϕ. We can obtain it as follows. First, write out a generalization of ϕ in prenex normal form, as x y Q ϑ, ∃ ∀ where Q is a string of quantifiers, and ϑ is quantifier-free. Introduce a new predicate R and form the sentence

x ( y (Q ϑ Rxy) y Rxy). ∃ ∀ → →∀ This has the second of the desired properties. It is also equivalent to

x ( y (Q ϑ Rxy) y Rxy), ∃ ∃ ∧¬ ∨∀ x y ((Q ϑ Rxy) z Rxz), ∃ ∃ ∧¬ ∨∀ x y (Q (ϑ Rxy) z Rxz), ∃ ∃ ∧¬ ∨∀ x y Q ((ϑ Rxy) z Rxz), ∃ ∃ ∧¬ ∨∀ x y Q z((ϑ Rxy) Rxz), ∃ ∃ ∀ ∧¬ ∨ This last sentence is in prenex normal form, though perhaps not be in Skolem normal form. Still, the number of universal quantifiers that pre- cede existential quantifiers has decreased. So the process terminates in a sentence that must be in Skolem normal form.

  Boolean rings and Stone spaces

. Boolean rings

In the ring (P(Ω), △, ), every element is its own square. As we said in §. on page , an arbitrary∩ ring with this property is called a Boolean ring. We now establish the promised Stone Representation Theorem [], that for every Boolean ring R the set Spec(R) of prime ideals of R is such that R embeds (as a ring) in (P(Spec(R)), △, ). This result is comparable to the so-called Cayley Theorem, that every∩ group G embeds in the group (Sym(G), , −1, id ) under g (x gx). ◦ G 7→ 7→ So let R be a Boolean ring, namely a ring that satisfies the identity

x2 = x.

We assume that R has a unit (like all other rings in these notes), but we need not assume that R is commutative, we shall prove it. First, R has characteristic 2, since

2x = (2x)2 = 4x2 = 4x, 0 = 2x, x = x. − It follows that R must be commutative, since now

x + y =(x + y)2 = x2 + xy + yx + y2 = x + xy + yx + y, 0= xy + yx, yx = xy = xy. − There are Boolean rings-without-units, for example Pω(ω) (the set of finite subsets of ω, defined on page ); but again, we shall not consider these as Boolean rings. We can rewrite the defining identity x2 = x for Boolean rings as

0= x2 x = x (x 1) = x (x + 1). − · − ·

 . Boolean rings

Hence if p is a prime ideal of R, then since 0 p, for all x in R we have either x p or x 1 p (that is, x + 1 p). Thus∈ p has just two cosets: ∈ − ∈ ∈ itself and 1+ p. Therefore R/p ∼= F2, a field, so p must be maximal. (We showed this for P(Ω) on page .) Also,

x p x + 1 / p. ∈ ⇐⇒ ∈

As above, we let Spec(R) be the set of prime ideals of R, and if x R, we let [x] be the set of prime ideals of R that do not contain x∈. If p Spec(R), we have ∈ xy p x p y p, ∈ ⇐⇒ ∈ ∨ ∈ hence xy / p x / p & y / p. ∈ ⇐⇒ ∈ ∈ Thus, [xy]=[x] [y]. ∩ Because R has characteristic 2, the sum of any two elements of x,y,x+y is the third. Since { }

xy(x + y)= x2 + xy2 = xy + xy = 0, p must contain at least one of x, y, and x + y. Then p contains exactly one of these, or all:

x + y / p (x p & y / p) (x / p & y p). ∈ ⇐⇒ ∈ ∈ ∨ ∈ ∈ Thus [x + y]=[x] △ [y].

Since finally [1] = Spec(R), we have now that x [x] is a homomorphism (of rings) from R to P(R). It is injective too,7→ since if x = 0, then x + 1 = 1, so x + 1 generates a proper ideal that does not contain6 x, and 6 this ideal is included in a prime ideal, which must be in [x], so [x] = ∅. 6

  Boolean rings and Stone spaces

. Ultrafilters

It follows now that an arbitrary Boolean ring R has all of the operations and relations that can be defined on a Boolean ring P(Ω) and its sub- rings. For example, R is (partially) ordered by 6 corresponding to , where ⊆ x 6 y xy = x. ⇐⇒ Also R has operations and , corresponding respectively to c and , so that ¬ ∧ ∩

x = x + 1, x y = xy. ¬ ∧ Then

0= x x, 1= 0, ∧¬ ¬ and the operation corresponding to can be given by ∨ ∪ x y = ( x y). ( ) ∨ ¬ ¬ ∧¬ ∗ The structure (R, , ) is called a Boolean algebra. Usually the signa- ture of Boolean algebras¬ ∧ is considered to be 0, 1, , , ; but again the additional operations can be defined from {and ¬. ∧ ∨} ¬ ∧ There is an equivalent axiomatic definition, which for the record can be given as follows. A structure (R, , ) is a Boolean algebra if and only if R has at least two elements, and¬ the∧ following equations are identities in the structure:

x y = y x, ∧ ∧ x (y z)=(x y) z, ∧ ∧ ∧ ∧ x x = x, ∧ x = x, ¬¬ (x x) y = x x, ∧¬ ∧ ∧¬ (x (y z)) = (x y) (x z). ¬ ∧¬ ∧ ¬ ∧¬ ∧¬ ∧¬ The last identity is equivalent to

x (y z)=(x y) (x z), ∧ ∨ ∧ ∨ ∧

 . Stone spaces when is defined as in ( ). The Boolean ring (R, +, , 0, 1) can be recov- ered from∨ the algebra by∗ the rules ·

x + y =(x y) ( x y), xy = x y. ∧¬ ∨ ¬ ∧ ∧ Now we have the characterization of ideals on page , as well as the definition of filters: A subset I of R is an ideal if and only if

0 I, ∈ x 6 y & y I = x I, ∈ ⇒ ∈ x I & y I = x y I, ∈ ∈ ⇒ ∨ ∈ and a subset F is a filter if and only if x +1: x F is an ideal, or equivalently { ∈ }

1 F, ∈ x F & x 6 y = y F, ∈ ⇒ ∈ x F & y F = x y F. ∈ ∈ ⇒ ∧ ∈ Then F is an ultrafilter if and only if x +1: x F is a prime ideal, and in this case F is the complement in{R of the prime∈ } ideal.

. Stone spaces

Given a Boolean ring R, we denote the set of its ultrafilters by

S(R).

We have a bijection p pc from Spec(R) to S(R); this induces an iso- morphism from P(Spec(7→ R)) to P(S(R)). In particular, if x R, we may now define ∈ [x]= f S(R): x f , { ∈ ∈ } so that x [x] is an embedding of R in P(S(R)). 7→ A reason for working with S(R) rather than Spec(R) is that, as we shall discuss in §. below, logical theories can be understood as filters, and then theories that are ultrafilters will be called complete theories. Also,

  Boolean rings and Stone spaces in a topological space X, the set of neighborhoods of a point is a filter of P(X). The set S(R) is called the Stone space of R. Indeed, since (as we saw)

[x] [y]=[xy], [1] = S(R), ∩ the subsets [x] of S(R) compose a basis for a topology on S(R) called the Stone topology. Every union of sets [x] is called is open in this topology, and it follows that: ) if is a finite collection of open sets, then is also open—here U U we may by convention allow ∅ = S(R); ) if is a collection of open sets, then Tis open—in particular, U U ∅ = ∅, so this is understoodT to be open. S The complementS of an open set is closed. Since

[x]c =[x + 1], the basic open sets [x] are also closed. Suppose f and g are distinct elements of S(R). Then we may assume frg has an element x. In this case

f [x], g [x + 1], [x] [x +1] = ∅. ∈ ∈ ∩ Thus the Stone topology is Hausdorff: it separates points.

Suppose x∈A[x]=S(R) for some subset A of R. Then every ultrafilter of R contains some element of A. Then no prime ideal of R includes A (since otherwiseS its complement would be an ultrafilter disjoint from A). Therefore the ideal (A) of R generated by A must the improper ideal R. In particular, this ideal contains 1. But then (A0) must contain 1 for some finite subset A0 of A. In this case no prime ideal of R includes A0, so every ultrafilter contains an element of A0, which means

[x]=S(R). x[∈A0 This shows the Stone topology is compact.

 . Stone spaces

Another way to prove this topology compact is the following. We have S(R) P(R), and we may identify the power 2R with P(R) under the map ⊆ (x : i R) i R: x = 1 i ∈ 7→ { ∈ i } from 2R to P(R). (This map can be understood as x x−1(1).) If we give the set 2 the discrete topology, then 2R can be7→ given the cor- responding product topology, which is the weakest topology in which all of the maps x xi (where i R) are continuous. Since the factors 2 are finite, this means7→ the topology∈ has a basis comprising, for each finite subset R of R, for each (a : i R ) in 2R0 , the set x 2R : (x : i 0 i ∈ 0 { ∈ i ∈ R0)=(ai : i R0) . This set is both open and closed. Then S(R) has the subspace∈ topology} and is moreover a closed subset of 2R, being the intersection of all closed sets of the forms

x 2R : x = 1 , { ∈ 1 } x 2R : x = 0 x = 1 , { ∈ ab ∨ b } x 2R : x = 0 x = 0 x = 1 , { ∈ a ∨ b ∨ ab } x 2R : x = 0 x = 1 . { ∈ a ⇐⇒ a+1 }

As 2R is compact by the Tychonoff Theorem, so must S(R) be compact.

Note that this case of the Tychonoff Theorem is easy to prove (and the general theorem is not much harder). Suppose is a collection of closed subsets of 2R with the finite-intersection property,F that is, for every finite subset 0 of , the intersection 0 is nonempty. We want to show that itselfF is nonempty.F We may assumeF that the elements of are basic closedF sets; and then we may assumeT that each element of hasF the form Tx 2R : x = e for some i in R and e in 2. By the finiteF intersection { ∈ i } property, there is no i in R such that contains both x: xi = 0 and x: x = 1 . Then contains a, whereF { } { i } F T e, if x: xi = e , ai = { }∈F (0, otherwise.

  Boolean rings and Stone spaces

We can compute [x] [y]=([x]c [y]c)c ∪ ∩ = ([x + 1] [y + 1])c ∩ = [(x + 1)(y + 1)]c =[xy + x + y + 1]c =[xy + x + y] =[x y]. ∨ Consequently, the only open sets that are also closed are the basic open sets, that is, the sets [x]. For if the open set x∈A[x] is also closed, then (since closed subsets of compact spaces are compact), we have S [x]=[x ] [x ]=[x x ] 0 ∪···∪ n−1 0 ∨···∨ n−1 x[∈A for some xi in A. Thus the subsets of S(R) that are both closed and open—clopen—compose a Boolean algebra or ring, which is the isomor- phic image of R under x [x]. 7→ An arbitrary compact Hausdorff space with a basis of clopen sets can be called a Stone space simply. Suppose S is one of these, and let B(S) be the set of clopen subsets of S. Then B(S) is a Boolean sub-ring of P(S). In the special case where S = S(R), we have just noted

R ∼= B(S(R)). In the general case, for every point P in S, the set U B(S): P U of basic neighborhoods of P is a filter and in fact an{ ultrafilter∈ of B(∈ S)}. Thus we have a map P U B(S): P U ( ) 7→ { ∈ ∈ } † from S to S(B(S)). Since S is Hausdorff, the map is injective. Suppose f S(B(S)). The intersection of every finite subset of f is an element of∈f; in particular, it is a nonempty subset of S. Thus f has the finite intersection property. Since S is compact, f itself must be a nonempty subset of S. Since also again S is Hausdorff, f must consist of a single point, P ; and then we must have T T f = U B(S): P U . { ∈ ∈ }

 . Boolean operations

So the map in ( ) is a bijection from S to S(B(S)). In fact it is a homeo- morphism. For,† it takes every clopen subset V of S (that is, every element V of B(S)) to the subset

U B(S): P U : P V { ∈ ∈ } ∈ of S(B(S)); and this subset is [V ], namely f S(B( S)): V f , since { ∈ ∈ } V U B(S): P U P V. ∈ { ∈ ∈ } ⇐⇒ ∈

. Boolean operations

We observed in §. that Boolean rings and Boolean algebras are ‘in- terdefinable’: every Boolean ring is also a Boolean algebra whose basic operations are the interpretations in the ring of certain terms; and then the Boolean ring can be obtained from the algebra in the same way. A Boolean operation on a power-set P(Ω) is just an operation on P(Ω) that is the interpretation of a term in the signature of rings or Boolean algebras. We should like to verify that every operation on P(Ω) that can be defined without reference to Ω itself is a Boolean operation. One way of doing this is as follows. Suppose we have n subsets X0, ..., Xn−1 of Ω. For each element σ of n 2 , there is a subset Xσ of Ω given by

X = X0 Xn−1, σ σ ∩···∩ σ where i i X , if σ(i) = 1, Xσ = i c ((X ) , if σ(i) = 0.

See Figure . for the cases n = 2 and n = 3 (here each set Xσ is labelled with σ). In case n = 0, the set 2n has the unique element 0 (the empty function), and then Xσ should be understood as Ω. In any case, the n n sets Xσ partition Ω into at most 2 subsets—or 2 subsets, if we still consider 2n as the set of functions from n to 2.| For| every subset S of n 2 in this sense, the subset σ∈S Xσ of Ω is a Boolean combination of the sets Xi; and every Boolean combination of these sets is of this form. S

  Boolean rings and Stone spaces

(0, 0, 0) (0, 0) (1, 0, 0)

(1, 1, 0) (1, 0, 1) (1, 0)(1, 1) (0, 1) (1, 1, 1)

(0, 1, 0)(0, 1, 1) (0, 0, 1)

Figure .: Boolean combinations

n Thus the number of Boolean combinations of the Xi is at most 22 . (It is less, if one of them is included in the union of the others.)

  More model theory

. The structure of definable relations

The structures that we have worked with so far are more precisely called one-sorted structures. By contrast, a vector-space is a two-sorted structure, with a sort for the vectors and a sort for the scalars. It is difficult to use a general notation for structures of arbitrarily many sorts; so we just describe the general situation for one-sorted structures, and then we point out that things can be adapted to many-sorted structures as needed.

The definable relations of a (one-sorted structure) A are themselves ele- ments of a many-sorted structure, with a sort for each finite subset I of ω: this sort corresponds to the set x : i I of variables. { i ∈ } The n-ary relations on A compose the set P(An). Suppose B A. The set of n-ary B-definable relations of A can be denoted by ⊆

n DefB(A).

This is a subset of P(An), and moreover it closed under the Boolean n operations; that is, DefB(A) is (the universe of) a Boolean sub-algebra of P(An). To check this, we need only note:

(ϕA)c =( ϕ)A, ϕA ψA =(ϕ ψ)A, ( ) ¬ ∩ ∧ ∗ and also A ∅ = xi = xi , 6 i

  More model theory

On the set P(An), if n > 1, additional operations besides the Boolean operations are possible. Indeed, if σ is a permutation of n, then there is a function σ∗ from An to An, given by

σ∗(x0,...,xn−1)=(xσ(0),...,xσ(n−1)).

Now the singulary operation X σ∗[X] on P(An) is induced. 7→ The same idea gives us functions from P(An) to P(Am). Indeed, sup- pose now σ is a function from m to n, where n> 0 or m = 0. For example, if m < n, then σ could be the inclusion of m in n. Or if m = n + 1, then σ could be given by

i, if i

σ∗(x0,...,xn−1)=(xσ(0),...,xσ(m−1)).

Then we have the function X σ∗[X] from P(An) to P(Am); here 7→ σ∗[X]= (xσ(0),...,xσ(m−1)): (x0,...,xn−1) X . { ∈ } For example, if n = m + 1, and σ is the inclusion of m in n, then

σ∗[X]= (x0,...,xm−1): (x0,...,xm) X ; { ∈ } but since the appearance of xm here may seem peculiar, we can also write

σ∗[X]= (x0,...,xm−1): (x0,...,xm) X for some xm . { ∈ } For another example, if m = 2 and n = 1, so that σ must be the constant function x 0 on 2, then 7→ σ∗[X]= (x,x): x X . { ∈ } m n In the general situation, we also have X σ∗[X] from P(A ) to P(A ), where 7→

σ [X]=(σ∗)−1[X]= (x0,...,xn−1) An : (xσ(0),...,xσ(m−1)) X . ∗ { ∈ ∈ }

 . Lindenbaum–Tarski algebras

For example, if σ is the inclusion of m in n, then σ∗[X] can be understood as X An−m; while if σ : 2 1, then × → σ [X]= x A: (x,x) X . ∗ { ∈ ∈ } Note that, in general, every function σ from m to n is a composition of:

. permutations of some k, . inclusions of some k in k + 1; . maps from some k + 1 to k as in ( ). † ∗ m n It should be clear that X σ [X] takes DefB (A) to DefB(A), and X σ [X] takes Defn (A) to7→Defm(A). 7→ ∗ B B The elements of ω are also finite subsets of ω. We could work with arbitrary finite subsets of ω. A formula whose free variables compose I the set xi : i I then defines in A a subset of A ; and then we have { ∈I } P I the subsets DefB(A) of (A ).

Consider the collection of all indexed families (PI : I Pω(ω)), where P P(AI ), and ∈ I ⊆

each PI is closed under the Boolean operations, • if σ : I J, the function X σ∗[X] takes P to P , and X • → 7→ J I 7→ σ∗[X] takes PI to PJ , for each n in ω, for each n-ary predicate R in S , P contains RA, • n for each n in ω, for each n-ary operation-symbol F in S , Pn+1 • contains (a,b) An+1 : F A(a)= b (that is, (F x = xn)A). for each c{in B, ∈P contains c . } • 1 { } I P ω The indexed family (DefB(A): I ω( )) is a minimum among these families. ∈

. Lindenbaum–Tarski algebras

For any signature S and n in ω, let us denote by

Snn(S )

  More model theory the set of n-ary formulas of S . Two elements ϕ and ψ of Snn(S ) are said to be logically equivalent if the formula ϕ ψ is a validity in the sense of Chapter , that is, for every structure A ↔of S ,

ϕA = ψA, or equivalently the sentence

x (ϕ(x) ψ(x)) ∀ ↔ is true in all structures in Mod(S ). The set of logical equivalence-classes of elements of Snn(S ) is denoted by

Bn(S ).

This is the nth Lindenbaum–Tarski algebra of S , and it is a Boolean algebra in a fairly obvious way. The rest of this section is devoted to spelling this out. The set Snn(S ) can be understood as a structure equipped with the operations and . A structure without relations, but only operations, is called an ¬algebra.∧ So we have an algebra (Snn(S ), , ). This algebra is not a Boolean algebra, simply because, for example,¬ ∧ ϕ is never the same formula as ϕ. ¬¬ Still, if A Mod(S ) and B A, the function ϕ ϕA is an epimorphism n∈S ⊆ n c 7→ from (Sn ( (B)), , ) to (DefB(A), , ): it is a surjection that respects the operations, as in¬ (∧ ) on page . ∩ ∗ For an arbitrary homomorphism h from an algebra M to N, the (gener- alized) kernel of h is the binary relation (x,y): h(x) = h(y) on M. If the structures are rings, then the connection{ between the usual} kernel and the generalized kernel is shown by the equivalence

h(x)= h(y) h(x y) = 0. ⇐⇒ − Such an equivalence is not always available for arbitrary algebras. Still, the generalized kernel is an equivalence-relation; and then a quotient M/ ker(h) can be defined, which is isomorphic to the image of M under h.

 . Theories and type-spaces

In particular then, there is a quotient of (Snn(S (B)), , ) that is a Boolean algebra, namely the quotient by the kernel of ¬ϕ ∧ ϕA: this n 7→ quotient is isomorphic to DefB(A). The kernel of a homomorphism on the arbitrary algebra M is an example of a congruence-relation: an equivalence-relation on M that respects the algebraic structure of M in the sense that, for every∼ n in ω, for every basic n-ary operation of M,

x0 y0 & & xn−1 yn−1 = F (x) F (y). ∼ ··· ∼ ⇒ ∼ In this case there is a well-defined quotient M/ , and there is a quotient- map from M to M/ whose kernel is of course∼ . ∼ ∼ We have seen the one-to-one correspondence between ideals and filters of a Boolean algebra. There is also a one-to-one correspondence between ideals and congruence-relations of a Boolean algebra, namely

I (x,y): x + I = y + I , 7→ { } with inverse x: x 0 . ∼ 7→ { ∼ } The set of congruence-relations on (Sn(S )S , , ) (or on any algebra) is closed under arbitrary intersections. The intersection¬ ∧ of the set of kernels of all homomorphisms ϕ ϕA, where A Mod(S ), is just the 7→ ∈ relation of logical equivalence defined above, and Bn(S ) is the quotient of Snn(S ) by this relation. We shall henceforth each formula with its logical equivalence-class, that is, confuse each element of Snn(S ) with its image in Bn(S ).

. Theories and type-spaces

By Gödel’s Completeness Theorem, the relation can be considered as ⊢ a binary relation on Bn(S ); indeed, it is the ordering induced by the algebraic structure. That is, ϕ ψ means that the formula ϕ ψ is a validity. With respect to the ordering,⊢ the greatest and least elements→ of the algebra (that is, the 1 and the 0) can be denoted respectively by

, . ⊤ ⊥

  More model theory

Every theory of S can be understood as a filter of B0(S ). That is, if Mod(S ), then Th( ) has the closure properties required of a filter:K ⊆ K ) it contains ; ) if it contains⊤ σ and τ, then it contains σ τ; ) if it contains σ, and if σ τ, then it contains∧ τ. ⊢ Suppose Γ B (S ). Then the set ⊆ 0 Th(Mod(Γ)) is a filter of B0(S ). This filter is the set of all elements of B0(S ) that are true in all models of Γ. But there is also a filter generated by Γ, namely the set of all σ such that Γ0 σ for some finite subset Γ0 of Γ. By the Compactness Theorem (Theorem⊢ , page ), these two filters are the same. If σ is a member, weV may write

Γ σ. ⊢

By the Compactness Theorem then, every proper filter of B0(S ) has a model. The sameness of the two foregoing filters is not trivial, because the Com- pactness Theorem is not trivial. The Compactness Theorem does not follow merely from the compactness of Stone spaces. For example, the second-order theory of (ω, 0,x x + 1) is axiomatized by 7→ xx + 1 = 0, ∀ 6 x y (x +1= y + 1 x = y), ∀ ∀ → X (0 X y (y X y + 1 X) y y X). ∀ ∈ ∧∀ ∈ → ∈ →∀ ∈ All models of these axioms are isomorphic to one another. In particular, there is no model that is also a model of each of the sentences c = 0, c = 1, c = 2, and so on, where c is a new constant. However, every6 finite6 set of6 these sentences has a model, if the interpretation of c is large enough. If T is a theory of S , then two formulas ϕ and ψ are equivalent modulo T if T x (ϕ(x) ψ(x)). ⊢∀ ↔

 . Saturation

The equivalence-classes with respect to this relation compose the Boolean algebra Bn(T ).

Then Bn(S ) is Bn(∅) or Bn( ). The Stone space of Bn(T ) can be denoted by ⊤ Sn(T ). The elements of this space are called n-types of T —or one may call them complete n-types of T , if one wants to use the word type more generally for arbitrary filters of Bn(T ) or just arbitrary collections of n- ary formulas. In any case, an element of S0(S ) is called a complete theory. Suppose Φ S (T ). If Φ is a finite subset of Φ, then Φ is not ∈ n 0 0 equivalent to modulo T . In this case T x Φ (x) has a model. ⊥ ∪ ∃ 0 V Suppose also A  T . An element a of An realizesn V Φ if,o for all ϕ in Φ,

A  ϕ(a).

In this case, we may say also that A itself realizes Φ. By Compactness, some model of T realizes Φ. Indeed, A has an elementary extension that realizes Φ.

. Saturation

We are often concerned with the parameters used in formulas and types. If M Mod(S ), and A is a subset of M, then by M we mean M, ∈ A considered as having signature S (A). We may then denote Sn(Th(MA)) simply by Sn(A); an element of this can be called a type over A. For every infinite cardinal κ, a structure is called κ-saturated if it realizes every type that has fewer than κ-many parameters. In particular, a structure is ω1-saturated or -saturated if it realizes all types in countably many parameters. ℵ1 Theorem . For every structure A with a countable signature, every ω non-principal ultrapower A /P of A is ω1-saturated.

  More model theory

Proof. If Φ is a type in countably many parameters, then Φ itself is countable, so we can write it as ϕn : n ω . Let an satisfy ϕ0 ϕn in A. Then { ∈ } ∧···∧ k 6 n = A  ϕ (a ). ⇒ k n Therefore, if P is a non-principal prime ideal of P(ω), then (an : n ω)/P realizes Φ in Aω/P . ∈

There is a version [, Thm .., p. ] of the foregoing for uncountable index-sets (or exponents) Ω; but then P must have a countable subset whose union is Ω (so one should show that such prime ideals can be found).

  Rings

. Ideals

In this chapter, as throughout these notes, the word ring will always mean a commutative unital ring. The letter R will always denote a ring in this sense. We have spent time with Boolean rings; now we work more generally (and also review some of the basic facts about integral domains and fields that we have already used). The main point is to develop some of the algebraic geometry that will be used in the examples of ultraproducts in the remaining two chapters. The following are equivalent conditions on the ring R: 1 = 0 in R. • R has only one element. • A ring meeting either of these conditions will be called trivial. Let the variables x and y range over R. If x = 0 and y = 0 (that is, if 6 6 x R r 0 and y R r 0 ), but ∈ { } ∈ { } xy = 0, then x and y are called zero-divisors. An integral domain is a non- trivial ring with no zero-divisors. Equivalently, R is an integral domain if and only if xy = 0 x =0 = y = 0. ∧ 6 ⇒ For example, the ring Z of rational integers is an integral domain. So are the polynomial rings K[X], K[X,Y ], and so forth, where K is a field.

Lang [, pp. –] recommends entire as the adjective form of integral domain, observing that integral would have been better, had it not already been taken for other purposes.

  Rings

However, suppose n is a positive integer. Then Z/nZ (namely the ring of integers modulo n) is an integral domain if and only if n is prime. Here nZ is the ideal nx: x Z of Z. In general, an ideal of R is an additive subgroup I such{ that∈ the} quotient group R/I is also a ring with respect to the multiplication given by

(x + I) (y + I)= xy + I. · That is, for I to be an ideal, we need

(x + I) I = I · (since I is the zero of R/I); this means

y I = xy I; ∈ ⇒ ∈ and this is enough. Every ideal contains 0 at least; if it contains nothing else, it is the trivial ideal or zero-ideal. A proper ideal of R is an ideal different from R itself: equivalently, it is an ideal that does not contain 1. A proper ideal I of R is called prime if

xy I & x / I = y I. ∈ ∈ ⇒ ∈ Compare this with the definition of integral domains: Theorem . A ring is an integral domain if and only if the trivial ideal is prime.

For another example, 0 zero-divisors is a prime ideal, which we shall denote by { } ∪ { } I0. So R is an integral domain if and only if I = 0 . A positive integer n 0 { } is prime if and only if nZ is a prime ideal of Z. Theorem . Let I be an ideal of R. Then R/I is an integral domain if and only if I is prime.

 . Ideals

Proof. We have that R/I is nontrivial if and only if I is proper. Assuming I is proper we have that the following are equivalent. R/I is an integral domain. • (x + I) (y + I)= I & x + I = I = y + I = I. • xy I ·& x / I = y I. 6 ⇒ • I is∈ prime. ∈ ⇒ ∈ • A unit of a ring is a divisor of 1. That is, if

xy = 1, then both x and y are units. The units of R compose a multiplicative subgroup of R, denoted by R×. In a trivial ring, 0 is a unit. but zero-divisors are never units or 0; that is, if R has any zero-divisors, they belong to R r (R× 0 ). We can also write ∪ { } 1 =0 = I R× = ∅. 6 ⇒ 0 ∩ A ring is a field if all nonzero elements are units. That is, R is a field if and only if R = R× 0 . In particular, fields have no zero-divisors, so they are integral domains.∪ { } The standard examples of fields are Q, R, and C. Moreover, if p is prime, then Z/pZ is (not only an integral domain, but) a field. A maximal ideal is a proper ideal that is maximal as such. An ideal I of R is maximal if I is a proper ideal and, for all ideals J of R,

I J & I = J = J = R. ⊆ 6 ⇒ If A R, then by ⊆ (A) is meant the smallest ideal of R that includes A. Then (A) is the set of sums a0x + + an−1x , 0 ··· n−1 where ai A and x R and n ω. In particular, n can be 0, and in ∈ i ∈ ∈ this case the sum above is 0. Thus (∅)= 0 , the trivial ideal. { }

  Rings

When A = a0,a1,... , we may write (A) as (a0,a1,... ). In particular, R = (1). Also{ the trivial} ideal is (0), where 0 is the ring element, not the empty set; but then this notation is redundant: strictly we should be able to write the zero-ideal as ( ). A proper ideal I of R is maximal if and only if, for all x in R r I, we have (I x )= R, ∪ { } that is, 1 (I x ), ∈ ∪ { } that is, for some y in R, 1 xy + I. ∈ Theorem . Let I be an ideal of R. Then R/I is a field if and only if I is maximal.

Proof. Again, R/I is nontrivial if and only if I is proper. Assuming I is proper we have that the following are equivalent. R/I is a field. • If x + I = I, then for some y, (x + I)(y + I)=1+ I. • If x / I,6 then for some y, 1 xy + I. • I is∈ maximal. ∈ • Corollary. Maximal ideals are prime.

The converse is true in Z.

. Localizations

Throughout this section, the ring R will be nontrivial. It will then have a quotient field (or field of fractions) constructed as Q is constructed from Z. A multiplicative subset of a ring is just a subset that is closed under multiplication. More precisely, if S is a multiplicative subset of R, this means that for all finite subsets ak : k < n of S, the product ak { } k

 . Localizations

k is also in S. In particular, 1 S, since k

For example, the set R r I0 of non-zero non-zero-divisors of R is a multi- plicative subset of R; more generally, so is the complement of any prime ideal; and so are 1 and R×. { } Lemma. If S is a multiplicative subset of R, there is an equivalence- relation on R S given by ∼ × (a,b) (x,y) (ay bx) s = 0 for some s in S. ( ) ∼ ⇐⇒ − · ∗

If R is an integral domain and 0 / S, or more generally if S I0 = ∅, then the equivalence-relation of the∈ lemma is given by ∩

(a,b) (x,y) ay bx = 0; ∼ ⇐⇒ − but we shall be interested in the more general situation, especially in Chapter ., and then a broader condition as in ( ) is needed. ∗ The -class of (a,b) is denoted by ∼ a b or a/b. The set of these classes is denoted by

S−1R; it can be called the localization of R at S.

Theorem . Let S be a multiplicative subset of R. The localization S−1R is a ring with respect to the usual operations:

a c ad + bc a c ac + = , = . b d bd b · d bd There is a homomorphism x x/1 from R to S−1R. This homomor- 7→ phism is injective if and only if S I = ∅. ∩ 0

  Rings

For the homomomorphism of the theorem, the condition of injectivity is not always met. It is met when 0 / S and also R is an integral domain ∈ (that is, I0 is trivial). In particular, we have Theorem . A ring is an integral domain if and only if it a sub-ring of a field.

Proof. For sufficiency, note that a zero-divisor of a sub-ring is a zero- divisor of the original ring. To show necessity, we note that an integral domain R embeds in the ring (R r 0 )−1R, which is a field. { }

The field (R r 0 )−1R in the proof is the quotient field (or field of fractions) of R{(assuming} R is a integral domain; otherwise this field is −1 (R r I0) R). Every substructure of a field is a ring. Therefore, by Theorem  on page , if T is field-theory, then T∀ is the theory of integral domains. The following is obvious, but should be noted:

Theorem . If an integral domain embeds in a field, then the embedding factors through the fraction field of the integral domain. That is, if R is an integral domain embedding under ι in its fraction-field K, and R embeds in a field L under ϕ, then K embeds in L under a map ψ such that ψ ι = ϕ. ◦

See Figure .. Suppose now K is a field, and L is a field of which K is

ϕ R / L ? ι ψ  K

Figure .: The universal property of the quotient field a subfield, that is, K L. In short, L/K is a field-extension. If n ω, ⊆ ∈

 . Localizations we may let X be an n-tuple (X0,...,Xn−1) of indeterminates, so that we can form the ring of polynomials in X over K, denoted by

K[X].

The fraction-field of this ring is denoted by

K(X); it is the field of rational functions in X over K. Suppose now a Ln. Then there is a function f f(a) (or Xi ai) from K[X] to L.∈ The range of this function is denoted7→ by 7→

K[a], and the fraction-field of this ring is denoted by

K(a); by the last theorem, we may consider this field as a subfield of L. In other words,

K[a] is the smallest sub-ring of L that includes K a0,...,an−1 ; • K(a) is the smallest subfield of L that includes K ∪ {a ,...,a }. • ∪ { 0 n−1} Note well that the function f f(a) is not generally defined on all of K(X), though it may be defined7→ on a sub-ring of this field that strictly includes K[X]. We shall consider this situation presently. Meanwhile, the kernel of the homomorphism f f(a) on K[X] is a prime ideal p, and then 7→ K[a] ∼= K[X]/p. If that ideal is nontrivial, then a is said to be algebraically dependent over K, or simply algebraic over K in case n = 1.

Theorem . If a is algebraic over K, then

K[a]= K(a).

Thus prime ideals of K[X] are maximal.

  Rings

Proof. If b + b a + + b an = 0, then 0 1 · ··· n · 1 b b b = 1 + 2 a + + n an−1 . a − b0 b0 · ··· b0 ·   Recall that Boolean rings also have the property that prime ideals are maximal (pages  and ). This is not generally true for K[X]. For example K[X,Y ]/(X Y ) ∼= K[X], an integral domain that is not a field; so (X Y ) is a non-maximal− prime ideal of K[X,Y ]. − In general, if S is a multiplicative subset of R, then, using the notation above, we can denote the localization S−1R also by R[S−1]. In case S is the complement of a prime ideal p of R, then this localization is denoted also by Rp. Confusingly, this may called the localization of R at p, although in the earlier terminology it is the localization of R at the complement of p. If R is an integral domain, then its fraction-field is the localization R{0}. Suppose again L/K is a field-extension and a Ln. If p is the prime ∈ ideal of K[X] that is the kernel of f f(a), then K[X]p is the largest sub-ring of K(X) on which the homomorphism7→ f f(a) is defined. 7→ In general, as we noted, if p is a prime ideal of R, then R r p is multi- plicative. The converse is true, and more, in the following sense: Theorem . Let I be a proper ideal of R. . R r I is multiplicative if and only if I is prime. . R r I = R× if and only if I is the unique maximal ideal of R.

A ring with a unique maximal ideal is called a local ring. Theorem . The localization of a ring at (the complement of) a prime ideal is a local ring, whose maximal ideal is generated by the image of that prime ideal.

We can now refer to Rp (where p is prime) as the local ring of R at p.A reason for the terminology can be seen in algebraic geometry, to which we now turn.

 . Algebraic geometry

Figure .: The zero-locus of y x2 in R −

. Algebraic geometry

Again suppose K and L are fields such that K L. For example, K ⊆ might be Q, and then L might be C. Let X be (X0,...,Xn−1) for some n in ω. Given a point x of Ln, we have looked at the homomorphism f f(x) from K[X] to L. ‘Dually’, an element f of K[X] determines a7→ function x f(x) from Ln to L. Often we are interested in the solution-set of7→ the equation

f(x) = 0.

Such equations are studied in school, at least when L = R and n = 1. We define Z (f)= x Ln : f(x) = 0 ; L { ∈ } this is the zero-locus of f in L. See Figure .. We may also want to look at more than one equation simultaneously, as for example in defining a straight line in R3. Accordingly, if A K[X], we define ⊆

ZL(A)= ZL(f), f\∈A that is, Z (A)= x Ln : f(x) = 0 ; ( ) L { ∈ } † f\∈A this is the zero-locus of A in L. See Figure .. (The definition should be compared with ( ) and ( ) on page .) The function A ZL(A) is the zero-locus map.∗ A† course in ‘analytic geometry’ is a7→ study of

  Rings

b

b

Figure .: The zero-locus of y x2,y x in R { − − } zero-loci in R, in case n is 2 or 3, where K[X] can be written as K[X,Y ] or K[X,Y,Z]. If I is an ideal of an arbitrary ring R, we define

√I = x R: xn I ; { ∈ ∈ } n∈ω [ this is the radical of I. (Note that the radical is indeed an ideal: if f n I and gm I, then (f + g)n+m−1 I.) An ideal is a radical ideal if it∈ is equal to∈ its own radical. ∈ Theorem . For all subsets A of K[X],

ZL(A) = ZL((A))=ZL(√(A)).

Thus all zero-loci are zero-loci of radical ideals.

The zero-loci of the various subsets (or just ideals, or just radical ideals) of K[X] are also called algebraic sets. As the notation is supposed to recall, the definition of ZL(A) depends on L. We intend to overcome this dependence. Meanwhile, we have the following. Theorem . The algebraic sets in L are the closed sets of a topology on Ln.

Proof. What this means, and what we shall show, is that () the inter- section of an arbitrary collection of algebraic sets is an algebraic set, and () the union of a finite collection of algebraic sets is an algebraic set. (In particular, since ∅ = ∅, this will be shown to be an algebraic set; S

 . Algebraic geometry also, ∅ is to be understood as Ln, so this too will be an algebraic set. See also page .) T The first point is obvious from the definition. Indeed, if is a collection of ideals of K[X], then A

Z (a) = Z ( ), L L A a \∈A X where is the ideal generated by . (So the ideal is ( ); but it consists ofA finite sums of elements of A, and the notation A is more suggestiveP of this.) In particular, as aS specialA case we have S A S P Ln = Z ( 0 ). L { }

For the second point, let a and b be ideals of K[X]. Then

a b = Z (a) Z (b). ⊆ ⇒ L ⊇ L Consequently Z (a b) Z (a) Z (b). L ∩ ⊇ L ∪ L Conversely, suppose x ZL(a b)rZL(a). Then for some f in a we have f(x) = 0. But then for∈ all g in∩b we have f g a b, so f(x) g(x) = 0, 6 · ∈ ∩ · and hence g(x) = 0 (since L is an integral domain). Thus x ZL(b). Therefore ∈ Z (a b) = Z (a) Z (b). ( ) L ∩ L ∪ L ‡ Finally

ZL((1)) = ∅. Thus finite unions of algebraic sets are algebraic.

The zero-locus of an arbitrary intersection of radical ideals need not be the union of the zero-loci of the ideals. For example, if K = Q and

k a = (X i) = (X 1) (X k) , k − − ··· − i=1 Y 

  Rings then Z (a )= 1,...,k , but a = 0 , so L k { } k∈N k { } T Z (a )= N L = Z ( 0 ) = Z ( a ). L k ⊂ L { } L k k[∈N k\∈N

The topology on Ln given by the theorem is called the Zariski topology over K. Then algebraic sets may be called Zariski-closed over K, or perhaps K-closed. All intersections of such sets are finite intersections, because of the following:

Theorem  (Hilbert Basis Theorem). For every n in ω, every ideal of the polynomial ring K[X0,...,Xn−1] is finitely generated.

Proof. The claim implies, and is therefore equivalent to, an apparently 0 n−1 stronger claim, namely that every ideal (A) of K[X ,...,X ] is (A0) for some finite subset A0 of A. For, if (A)=(f0,...,fm−1), then each k k fk is in (A ) for some finite subset A of A; and then we can let A0 = k k

SThe claim as also equivalent to the claim that every sequence (ak : k ω) of ideals such that ∈ a a a 0 ⊆ 1 ⊆ 2 ⊆··· —that is, every increasing chain of ideals (indexed by ω)—is eventually constant. For, the union of such a chain is an ideal b, and if this ideal is finitely generated, then it has a generating set whose elements all lie in some aℓ, and then this ideal is b. Conversely (or inversely), if a were not finitely generated, then for all subsets f : k <ℓ of a we could { k } find fℓ in a r (fk : k<ℓ); thus we could form a strictly increasing chain ((f : k<ℓ): ℓ ω). k ∈ There is now also a fourth form of our claim: every countably generated ideal is finitely generated. We turn to proving the claim, in any convenient form. The claim is trivially true when n = 0, since a field has only two ideals: the trivial ideal and the improper ideal (1). The claim is still easy when n = 1, because K[X] is a Euclidean domain. In particular, if f and g are in K[X], we have an algorithm (the Euclidean

 . Algebraic geometry algorithm) for finding their greatest common divisor—say h; and then (f,g)=(h). Hence if a =(fk : k ω), for each k in ω we can find gk so that ∈ (f0,...,fk)=(gk). In particular, g divides g . Then min deg(g ): k ω = deg(g ) for k+1 k { k ∈ } ℓ some ℓ, and consequently a =(gℓ). When n > 2, we have not got the Euclidean algorithm; but we can come close enough if we use induction. Suppose then that the claim is true when n = m. Let a be an ideal of K[X0,...,Xm]. We form a sequence (f0,f1,... ) of elements of a by recursion: Given (fk : k<ℓ), we let fℓ, if it exists, be an element of a r (fk : k <ℓ) of minimal degree as a polynomial in Xm over K[X0,...,Xm−1]. Then these degrees form an increasing sequence:

degXm (f0) 6 degXm (f1) 6 degXm (f2) 6 ...

m Let gk be the leading coefficient of fk (as a polynomial in X over 0 m−1 0 m−1 K[X ,...,X ]; so gk K[X ,...,X ]). By inductive hypothe- sis, for some ℓ, ∈ (g : k ω)=(g : k<ℓ). k ∈ k Then in particular g (g : k<ℓ). ℓ ∈ k Therefore (fk : k <ℓ) has an element h whose leading coefficient (as m 0 m−1 a polynomial in X over K[X ,...,X ]) is gℓ; and the degree of this element can have any value that is at least the degree of fℓ−1. In particular we may assume h has the degree of f . But then f h has ℓ ℓ − lower degree and belongs to a r (fk : k<ℓ); that is, fℓ did not have minimal degree. Thus there is no fℓ; that is, a =(fk : k<ℓ).

A singly generated ideal is called principal. Then part of our proof of the theorem gives the following:

Porism. Every ideal of K[X] is principal.

Hence, although in the example above N is the union of zero-loci, it cannot itself be a zero-locus; for, every zero-locus of polynomials in one

  Rings variable is the zero-locus of a single polynomial, so it is either the whole field L or a finite subset of this.

The theorem itself has the following:

Corollary. Every decreasing chain of closed subsets of Ln is eventually constant.

It is not obvious that the corollary implies the theorem, since it is not obvious that a b implies Z (a) Z (b), even if a and b are radical ⊂ L ⊃ L ideals. In fact, this can be false. For example, when L = R, then (X2 +1) is a radical ideal whose zero-locus is the same as the zero-locus of (1), namely empty.

. A Galois correspondence

For a ‘dual’ to the definition of zero-loci, if A Ln, we define ⊆

IK (A)= f K[X]: f(x) = 0 ; x { ∈ } \∈A this is a radical ideal of K[X], and we may call it the ideal of A over K. The function A IK (A) is the ideal map. The definition does not involve the field 7→L as such, but it does involve K, as the notation indicates.

Every subset A of Ln has a Zariski closure over K, namely the smallest K-closed subset of Ln that includes A. We can denote this closure by

K A .

Then we have easily:

Theorem . For all subsets A of Ln,

K K A = ZL(IK (A)), IK (A)=IK (A ).

 . A Galois correspondence

In the proof of Theorem  we used that the zero-locus map is inclusion- reversing. This is by the form of the definition. Since the definition of the ideal map has the same form, this function too is inclusion-reversing. The last theorem theorem is entirely formal in this way. Indeed, the form of the definitions of the zero-locus and ideal maps is just the form used in establishing the original Galois correspondence in field theory. In that context, L is a normal and separable finite extension of K. We let G be the group of automorphisms of L over K, that is, G = Aut(L/K). For subgroups H of G, we define

Fix(H)= x L: xσ = x ; { ∈ } σ\∈H for intermediate extensions F of L/K, we have

Aut(L/F )= σ G: xσ = x . { ∈ } x\∈F If we abbreviate Fix(H) as H′, and Aut(L/F ) as F ′, then we always have

X Y = X′ Y ′, X X′′, ⊆ ⇒ ⊇ ⊆ so that (as a special case of the latter) X′ X′′′, but also X′ X′′′. Thus ⊆ ⊇ X′′′ = X′. Therefore we have a one-to-one correspondence X X′ between the fields Fix(H) and the groups Aut(L/F ); and this is entirely7→ by the general form of the definitions. What makes the correspondence useful is the special features of the situation: that G =[L : K], simply because L/K is normal, separable, and finite; and| then| Aut(L/F ) = [L : F ] for all intermediate extensions F of L/K; and therefore| every|intermediate field of L/K is Fix(H) for some H, and every subgroup of G is Aut(L/F ) for some intermediate field F of L/K. We return to the algebraic-geometric situation. For the formal reasons just discussed, we have a one-to-one Galois correspondence between the K-closed subsets of Ln and certain radical ideals of K[X], namely those of the form IK (F ) for some K-closed set F . See Figure .. We

  Rings

(x,y) (x 1,y 1) ● − − ●● rr ●● rrr ●● rr b ● rrr (y x,y x2) − −❑ ①① ❑❑ b ① ❑ ①① ❑❑❑ ①① ❑❑ ①① ❑ (y x) (y x2) − − Figure .: Algebraic-geometric Galois correspondence may refer to those ideals as being themselves L-closed. Is this Galois correspondence of any use?

The L-closed ideals of K[X] are just those of the form IK (ZL(a)) for some radical ideal a of K[X]; and then

a I (Z (a)). ( ) ⊆ K L § This is an equation if and only if a is L-closed. We noted in effect that if L = R then the radical ideal (X2 + 1) is not L-closed:

(X2 + 1) (1) = I (Z ((X2 + 1))). ⊂ K R

However, as L grows larger, so does ZL(a); but then IK (ZL(a)) becomes smaller. In fact 2 2 (X +1)=IK (ZC((X + 1))). We now are faced with the following:

Question . For every radical ideal a of K[X], is there an extension L of K large enough that

a =IK (ZL(a))?

Question . Is there an extension L of K large enough that for all ideals a of K[X] and all extensions M of K,

I (Z (a)) I (Z (a))? K L ⊆ K M

 . A Galois correspondence

Note well that a and L are quantified in different orders in the two ques- tions, as a L and L a respectively. This means a positive answer to Question∀ ∃  does not∃ immediately∀ give a positive answer to Question . We would first have to show that the different fields L corresponding to the different ideals a are all included in one large field. This is true however, since the class of fields has the joint embedding property: If f0 embeds K in L0, and f1 embeds K in L1, then there is a field M, and there are embeddings gi of the Li (respectively) in M, such that g f = g f . See Figure .. 0 ◦ 0 1 ◦ 1

M ❇ g ⑤⑤ ❇❇ g 0 ⑤⑤ ❇❇ 1 ⑤⑤ ❇❇ ⑤⑤ ❇ L0 L1 ❇❇ ⑤ ❇❇ ⑤⑤ f ❇❇ ⑤⑤f 0 ❇ ⑤⑤ 1 K

Figure .: Joint embedding property of fields

By contrast, even if Question  has a positive answer, it is not at all clear that the answer to Question  must be positive. We settle Question  first in a special case.

Lemma. For all maximal ideals m of K[X], for all extensions L of K in which K[X]/m embeds over K,

m =IK (ZL(m)).

Proof. As formulated here, the lemma almost proves itself. We just have to show IK (ZL(m)) is a proper ideal. But the image of X in K[X]/m is in the zero-locus of m. In particular, if L includes this field, then ZL(m) is not empty, so IK (ZL(m)) cannot be all of K[X].

This gives us another special case:

  Rings

Theorem . If K[X]alg L, for all ideals a of K[X] such that ⊆ IK (ZL(a)) is the improper ideal,

a =IK (ZL(a)).

Proof. The claim is

I (Z (a))=(1) = a = (1). K L ⇒ We prove the contrapositive. If a is a proper ideal of K[X], then it is included in some maximal ideal m. The field K[X]/m can be understood as an algebraic extension of K(Xi : i I) for some subset I of n, so it alg ∈ embeds in K(X) . By the lemma then, since IK (ZL(m)) is a proper ideal, so is IK (ZL(a)).

Note that if IK (ZL(a)) = (1), then ZL(a) = ∅. Thus every proper ideal has non-empty zero-locus6 in a large-enough6 field. Nullstellensatz means zero-locus theorem:

Theorem  (Nullstellensatz). If K[X,Y ]alg L, for all radical ideals a of K[X], ⊆ a =IK (ZL(a)).

Proof. Say f I (Z (a)). If x Z (a), then f(x) = 0. This shows ∈ K L ∈ L Z (a f 1 )= ∅, so L ∪ { − } I (Z (a f 1 )) = (1). K L ∪ { − } By the last theorem, a f 1 too must generate the improper ideal of K[X]. We want to be∪ able { − to} conclude f a. To do so, we modify the ∈ argument so far. We have f Y IK (ZL(a)), if we consider a now as a subset of K[X,Y ]. As before,· a ∈ f Y 1 must generate the improper ideal of K[X,Y ]. Now, by itself,∪a { generates· − } the ideal of K[X,Y ] whose elements are polynomials in Y with coefficients from a. Hence there is some such polynomial g, and there is some h in K[X,Y ], such that

g + h (f Y 1) = 1. · · −

 . A Galois correspondence

Substituting 1/f for Y , we get g(1/f) = 1; that is, 1 1 g + g + ...g = 1 0 1 · f m · f m for some gi in a, and hence

g f m + g f m−1 + + g = f m. 0 · 1 · ··· m This means f m a. Assuming a is radical, we have f a. Thus I (Z (a)) a and∈ therefore I (Z (a)) = a. ∈ K L ⊆ K L We have now settled both Questions  and . This suggests that under- standing algebraic sets can somehow be reduced to understanding radical ideals of K[X]. Indeed, there is some extension L of K large enough that we have a Galois correspondence between the K-closed subsets of Ln and the radical ideals of K[X]. It is not particularly important for what fol- lows that this field L can be chosen as Kalg. Nonetheless, it is true:

Theorem  (Hilbert’s Nullstellensatz, weak form). Every proper ideal of K[X] has a non-empty zero-locus in every extension of Kalg.

Proof. In the lemma, by the Hilbert Basis Theorem, m has the form (f0,...,fℓ) for some fi in K[X]. Thus the formula

f = 0 f = 0 0 ∧···∧ ℓ has a solution in K[X]/m and a fortiori in (K[X]/m)alg. The latter field is an elementary extension of Kalg, by the model-completeness of the theory of algebraically closed fields (Theorem  on page ). Therefore the formula has a solution here too. Thus as long as Kalg L, we have Z (m) = 0. ⊆ L 6 Now the proof of Theorem  gives:

Corollary (Hilbert’s Nullstellensatz, strong form). For all radical ideals a of K[X], IK (ZKalg (a)) = a.

  Finite fields

. Ultraproducts of finite structures

Suppose a theory T has arbitrarily large finite models. Then there is a sequence (Am : m ω) of finite models of T such that Am >m in each case. Consequently,∈ the sentence | |

(x ,...,x ) x = x ∃ 0 m i 6 j i

. Finite fields

Let us review the basic theorems about finite fields. Suppose K is a field. There is a homomorphism 1 1 (or k k 1) from Z to K. The kernel of 7→ 7→ · this homomorphism is nZ for some positive n, called the characteristic of K, char(K). Since Z/nZ must be an integral domain (by Theorem , p. ), n is either 0 or prime. If char(K) = 0, we may consider Q as a subfield of K; if char(K) is a prime p, we consider Z/pZ, denoted by Fp, as a subfield of K. Respectively, Q or Fp is the prime field of K. Let K be a finite field of characteristic p. Then K is a vector-space over m × Fp of some finite dimension m, so K has order p . The group K of

 . Finite fields

m units of K has order pm 1, so its every element is a root of xp −1 1. Then every element of K−is a root of the polynomial −

m xp x. − Since the formal derivative of this is 1, it has no repeated roots. Thus alg − its roots (in an algebraic closure Fp of Fp that includes K) are precisely the elements of K: we have

m K = x F alg : xp = x . { ∈ p } m Conversely, for all m in N, since the map x xp is an automorphism alg alg pm 7→ of Fp , the set x Fp : x = x (namely the fixed field of the automorphism) is{ a subfield∈ having order} pm. This then is the unique alg subfield of Fp of this order, and we can denote it by

Fpm .

× The group Fpm of units of this field is cyclic. For again, it is a finite abelian group of order pm 1 and is therefore a direct product −

Gℓ, m ℓ|pY−1 where each Gℓ is an ℓ-group (a group whose elements have orders that are powers of ℓ; here and elsewhere in this chapter, ℓ is, like p, a prime number). Since Gℓ is finite, for some positive integer n, every element of Gℓ is a solution of n xℓ = 1. But in a field, this equation has no more than ℓn solutions. Therefore, if n × n is minimal, Gℓ must be cyclic of order ℓ . Then the product Fpm is itself cyclic, of order pm 1. − alg The collection of finite subfields of Fp , ordered by inclusion, is iso- morphic, under the map Fpm m, to N as ordered by dividing. That is, 7→ F m F n m n. p ⊆ p ⇐⇒ | See Figure .. Indeed, if F m F n , then F n is a vector-space over F m , p ⊆ p p p

  Finite fields

Fp8 Fp12 Fp18 Fp27 ❇ ❈ ❈ ❇❇ ④④ ❈❈ ④④ ❈❈ ④④ ❇❇ ④④ ❈❈ ④④ ❈❈ ④④ ❇❇ ④④ ❈❈ ④④ ❈❈ ④④ ❇ ④④ ❈ ④④ ❈ ④④ Fp4 Fp6 Fp9 ❈ ❈ ❈❈ ④④ ❈❈ ④④ ❈❈ ④④ ❈❈ ④④ ❈❈ ④④ ❈❈ ④④ ❈ ④④ ❈ ④④ Fp2 Fp3 ❈ ❈❈ ④④ ❈❈ ④④ ❈❈ ④④ ❈ ④④ Fp

Figure .: The lattice of finite fields of characteristic p so its order is (pm)k for some k, and then n = mk, so m n. Conversely, if m n, then | | pm 1 pn 1, − | − and therefore m n xp −1 1 xp −1 1, − | − so F m F n . p ⊆ p Finally, alg F = F n ( ) p p ∗ n[∈N (since every extension Fpn /Fp is certainly algebraic, while every finite algebraic extension of Fp is a finite field).

. Galois groups

We have shown that for each prime p, for each m in N, there is a subfield alg Fpm of Fp , and this subfield is generated by (in fact it consists of) m the roots of the polynomial xp x, which is separable. Therefore the − finite field-extension Fpm /Fp is normal and separable, that is, Galois.

 . Galois groups

The order of its group of automorphisms is [Fpm : Fp], that is, m. But the Frobenius automorphism of F alg, namely x xp or p 7→ Frob, restricts to an automorphism of Fpm of order m, since we have shown in effect k Fix(Frob )= Fpk . Thus Aut(F m /F )= Frob ↾ F m = Z/mZ. p p h p i ∼ For any field K, let us write Gal(K) = Aut(Ksep/K), the absolute Galois group of K. We want to determine Gal(Fp). Suppose σ Gal(F ). For every n in N, we have ∈ p σ ↾ F n Aut(F n /F ), p ∈ p p and hence for some σ(n) in Z

σ(n) σ ↾ Fpn = (Frob ↾ Fpn ) . All that matters here is the congruence-class of σ(n) modulo n. Thus we have an injective map σ (σ(n): n N) 7→ ∈ from Gal(F ) to Z/nZ. The map is not surjective, but if m n, p n∈N | then since Fpm Fpn we must have ⊆Q σ(n) σ(m) (mod m). ≡ However, suppose an element (σ(n): n N) of Z/nZ meets this ∈ n∈N condition. For any x in F alg we can define an element σ of Gal(F ) by p Q p σ(m) xσ = xp ,

σ where x Fpm . (Here x is of course the image of x under σ.) This ∈ σ definition of x is independent of the choice of m, since if also x Fpn , then ∈ x F gcd(m,n) , ∈ p

  Finite fields so σ(m) σ(gcd(m, n)) σ(n) (mod gcd(m, n)) ≡ ≡ and therefore σ(m) σ(gcd(m,n)) σ(n) xp = xp = xp . Thus

Gal(F ) = (σ(n): n N) Z/nZ: πn (σ(n)) = σ(m) p ∼ { ∈ ∈ m } iY∈N m^|n where πn is the quotient-map x + nZ x + mZ from Z/nZ to Z/mZ. m 7→ In particular, Gal(Fp) has a certain ‘universal property’ with respect to n the system of groups Z/nZ and homomorphisms πm: G . Gal(Fp) is a group G from which there is a homomorphism hn to Z/nZ for every n in N such that, if m n, then | πn hG = hG . m ◦ n m . For every such group G, there is a unique homomorphism h from G to Gal(Fp) such that, for each n in N,

hG = hGal(Fp) h. n n ◦

See Figure .. Therefore Gal(Fp) is called a limit of the given system of

Gal(Fp) O ✽ ✝✝ ✽✽ ✝✝ h ✽✽ F ✝ ✽ F Gal( p) ✝ ✽ Gal( p) hn ✝✝ G ✽✽hm ✝ t ❑❑ ✽ ✝✝ tt ❑❑ ✽✽ ✝✝ tt ❑❑ ✽✽ tt G G ❑❑ ✽ ✝✝ t h h ❑ ✽ Ó✝ ytt n m ❑%  Z/nZ n / Z/mZ πm

Figure .: The universal property of Gal(Fp) groups and homomorphisms. This is the category-theoretic sense of limit

 . Pseudo-finite fields as given in, say, [, p. ] or []. Every set of groups, equipped with some homomorphisms, has a limit in this sense, though the limit might be empty.

The group Gal(Fp) is called more precisely a projective limit or an inverse limit of the system of groups Z/nZ with the quotient-maps, because any two of these groups are quotients of a third. This condition is not required for the existence of the limit. We give the finite groups Z/nZ the discrete topology, and their product the product topology. This product is compact by the Tychonoff Theorem (mentioned above on page ). The image of Gal(Fp) in this group is closed, so it too is compact: it is called a pro-finite completion of the system of finite cyclic groups.

. Pseudo-finite fields

Two examples of infinite models of the theory of finite fields are:

F /M, F n /M, ( ) p p † prime p Y nY∈N where in each case M is some non-principal maximal ideal. The first example has characteristic 0; the second, characteristic p. By the ‘Riemann Hypothesis for curves’ as proved by Weil, for every prime power q, for every curve C of genus g over Fq, the number of Fq-rational points of C is at least

1+ q 2g√q. − In particular, if q is large enough, then C does have an Fq-rational point. A field K is called pseudo-algebraically-closed or PAC if every plane curve defined over K has a K-rational point. This condition entails that every absolutely irreducible variety over K has a K-rational point.

Perhaps one should talk about convergent sequences here. . . See for example [, Ex. V.., p. ] or [, Thm ., p. ]. See [, ch. , pp. –].

  Finite fields

The following are now true of every infinite model of the theory of finite fields: . It is perfect. . It has exactly one extension of each degree (in some algebraic clo- sure). . It is pseudo-algebraically-closed. This is not obvious, even given the results stated above; one must show that these conditions are first-order, that is, the structures that satisfy them make up an elementary class. By the definition of Ax [], a field with the first two of these properties is quasi-finite; with all three of these properties, pseudo-finite. So every infinite model of the theory of finite fields is (quasi-finite and) pseudo-finite. Ax proves the converse. In particular, Ax proves that every pseudo-finite field is elementarily equivalent to a non-principal ultraproduct of finite fields, and indeed to one of the ultraproducts given above in ( ). The method is as follows; here I use Ax [] and also Chatzidakis [].† For every field K, the field Abs(K) of absolute numbers of K consists of the algebraic elements of K (here algebraic means algebraic over the prime field). The following is [, Prop. ′, §, p. ].

Lemma. For every field K of prime characteristic p, there is a maximal n ideal M of n∈N Fp such that Q Abs(K) ∼= Abs( Fpn /M). nY∈N

alg Proof. Because F = F n as in ( ) on page , we need only p n∈N p ∗ choose M so that, for all m in N, S

F m K F m F n /M. p ⊆ ⇐⇒ p ⊆ p nY∈N

For each m in N, let fm be an irreducible element of Fp[X] of degree m. Then each zero of fm generates Fpm over Fp. So we want M to be such that F m K f has a zero in F n /M. p ⊆ ⇐⇒ m p nY∈N

 . Pseudo-finite fields

Let F be the ultrafilter on N corresponding to M, that is, F = Nr supp(f): f M = n: f = 0 : f M . { ∈ } { n } ∈ } Then 

f has a zero in F n /M n: f has a zero in F n F. m p ⇐⇒ { m p ∈ n∈N Y Moreover, f has a zero in F n m n. m p ⇐⇒ | So, combining all of our equivalences, we want to choose F on N such that F m K n: m n F. p ⊆ ⇐⇒ { | }∈ For each m in N, the subset

k : k m & F k K { | p ⊆ } of N is a sublattice of the lattice of factors of m with respect to divisibility: in particular, it contains the least common multiple of any two members. It also contains 1. Therefore it has a maximum element, say g(m). The arithmetic function g is multiplicative: gcd(m, n)=1 = g(mn)= g(m) g(n). ⇒ · Now let b = x: gcd(m,x)= g(m) . m { } Then the function m b is also multiplicative, in the sense that 7→ m gcd(m, n)=1 = b = b b . ( ) ⇒ mn m ∩ n ‡ Indeed, suppose gcd(m, n) = 1. Then or all x in N, gcd(mn,x) = gcd(m,x) gcd(n, x), · and these factors are co-prime, being respectively factors of m and n. But also g(mn) = g(m) g(n), and these factors are co-prime, being respectively factors of m and· n. Therefore gcd(mn,x)= g(mn) gcd(m,x)= g(m) & gcd(n, x)= g(n). ⇐⇒ Thus it contains the least common multiple of every (finite) set of members, includ- ing the empty set.

  Finite fields

So we have ( ). Moreover, we have also ‡

m 6 n = b n b m . ( ) ⇒ ℓ ⊆ ℓ § For, we have

g(ℓn) y : ℓ ∤ y , if g(ℓn) <ℓn, bℓn = { · } ℓny : y N , if g(ℓn)= ℓn, ({ ∈ } and also m 6 n = g(ℓm) = min ℓm,g(ℓn) . ⇒ Now we can just check that ( ) holds in each of the three
cases § g(ℓn)= ℓn, ℓm

So we have finally b b = b . m ∩ n lcm(m,n) Thus, since each bm is nonempty, the set of these generates a proper filter on N. Let F be an ultrafilter on N that contains all of the sets bm. We claim that this F is as desired. Indeed,

if F m K, so g(m)= m, then b = mx: x N ; • p ⊆ m { ∈ } if F m * K, so g(m)

F m K, p ⊆ mx: x N F, { ∈ }∈ fm has a root in Fpn /M, nY∈N F m F n /M. p ⊆ p nY∈N

The lemma has a companion [, Prop. ], namely that for every quasi-

finite field K of characteristic 0, there is a maximal ideal M of p Fp such that Q Abs(K) = Abs( Fp/M), p Y

 . Pseudo-finite fields but the proof is more difficult. Since all fields of characteristic 0 are per- fect, quasi-finiteness in this case just means having exactly one extension of each degree. In this case the field of absolute numbers has at most one extension of each degree. This is because, if α is algebraic over Abs(K), then α has the same degree over K that it has over Abs(K). For, the minimal polynomial of α over Abs(K) is a product

(X α ), − i i

Abs(F ) = Abs(F ′) = F F ′. ( ) ∼ ⇒ ≡ ¶ With this and the foregoing lemma, we shall have that every pseudo-finite field (at least in positive characteristic) is elementarily equivalent to an ultraproduct of finite fields. To establish ( ), since Abs(F ) is determined by Th(F ), we can replace F and F ′ (respectively)¶ by elementarily equivalent fields. In particular, we can replace them with ultrapowers with exponent ω; these ultrapowers are ω1-saturated by Theorem  on page . Now take a countable ele- mentary substructure F0 of F ; this exists by the downward Löwenheim– Skolem–Tarski Theorem, Theorem . One shows [, ., Lemme de ′ ′ plongement] that this embeds in F under a monomorphism ϕ0. Then F ′ ′ has an elementary substructure F0 that includes the image of F0; and F0 ′ −1 embeds in F under a monomorphism ϕ0 that extends ϕ0 . Continuing, ′ ′ we obtain isomorphic elementary substructures Fω and Fω of F and F respectively. See Figure .. This establishes ( ). ¶

  Finite fields

F F ′

S ϕ n∈ω n ′ Fω / Fω

ϕ1 F1 / ϕ1[F1]

ϕ′ ′ ′ 0 ′ ϕ0[F0] o F0

ϕ0 F0 / ϕ0[F0]

Figure .: Isomorphisms of pseudo-finite fields

  Schemes

Sources for the algebraic geometry of this chapter include Coombes [] and Hartshorne []. The main point is to look at the ultraproduct scheme at the end; this work is based on the first of the three MSRI/Evans Hall Lectures, given at the University of California at Berkeley in the spring of  by Angus Macintyre.

. The spectrum of a polynomial ring

In §., letting f range over K[X] where X = (X0,...,Xn−1), and letting x range over Ln where K L, we used the equation f(x) = 0 to establish a Galois correspondence⊆ between the K-closed subsets of Ln and certain radical ideals of K[X]. If L is large enough (by Theorem  on page )—more precisely, if L includes Kalg (by Theorem  and its corollary)—, then the Galois correspondence is between the K-closed subsets of Ln and (all of) the radical ideals of K[X]. In particular, the correspondence is inclusion-reversing. Thus the set of radical ideals of K[X] encodes the topological structure of Ln for sufficiently large L, even for L including Kalg.

Suppose indeed Kalg L. We want more than the Galois correspon- dence. Suppose we are⊆ given a particular f in K[X]. We are interested in its zero-locus, the K-closed set ZL(f); and this now corresponds to IK (ZL(f)), which is √(f). We should should like to have a way of pick- ing out this ideal among all of the radical ideals of K[X], without having to refer to Ln. One way of doing this is simply to observe that √(f) is the intersection of all radical ideals of K[X] that contain f. We shall develop an alternative that is in some ways more satisfying.

These lectures used to be preserved on the MSRI website; but I could not find them there, the last time I looked.

  Schemes

If a Ln, let us write I (a) for I ( a ). That is, ∈ K K { } I (a)= f K[X]: f(a) = 0 . K { ∈ } This is always a prime ideal. In particular, it is radical. Moreover, for every subset A of Ln,

IK (A)= IK (x). x \∈A In particular,

IK (ZL(f)) = IK (x). x ∈\ZL(f) Thus IK (ZL(f)) is the intersection of some prime ideals of K[X] that include it. Therefore it is the intersection of all prime ideals of K[X] that include it. Let us henceforth denote K[X] simply by R. Then the set of all prime ideals of K[X] is denoted by

Spec(R).

(This will later be understood as part of the spectrum of R.) Usually p will be understood as ranging over this set. We have the following.

Theorem . For all radical ideals a of R,

a = p Spec(R): a p . { ∈ ⊆ } \ In particular, for all f in R,

√(f)= p Spec(R): f p . { ∈ ∈ } \ Proof. If L is large enough, then we now have

a =I (Z (a)) = p: I (Z (a)) p = p: a p . K L { K L ⊆ } { ⊆ } \ \ Also, for all p in Spec(R),

√(f) p f p. ⊆ ⇐⇒ ∈

 . The spectrum of a polynomial ring

This theorem is true for arbitrary rings R (see Theorem  below); but in the present case the work of proving it has already been done for the Nullstellensatz (Theorem ). In the theorem, the condition that f p is equivalent to the condition that f + p = 0 in R/p. Moreover: ∈ Theorem . The homomorphism

f (f + p: p Spec(R)) 7→ ∈ from R into the product R/p p∈Spec(Y R) is an embedding.

Proof. The kernel of this map is Spec(R), which is the trivial ideal since this ideal is prime. T So if L is large enough, that is L Kalg, we have a one-to-one corre- spondence between: ⊇ n closed subsets ZL(a) of L ; • radical ideals a of R; • subsets p Spec(R): a p of Spec(R). • { ∈ ⊆ } Moreover, suppose we write

(fp : p Spec(R))=(f + p: p Spec(R)). ∈ ∈ That is, we consider this as a function p fp on Spec(R), where fp R/p. If A R, then 7→ ∈ ⊆ p Spec(R): A p = p Spec(R): f p { ∈ ⊆ } { ∈ ∈ } f\∈A = p Spec(R): fp = 0 . { ∈ } f\∈A This is a kind of zero-locus. If f R, we can define ∈ Z(f)= p Spec(R): fp = 0 { ∈ }

  Schemes

(with no subscript on the Z this time), and if A R, we define ⊆ Z(A)= Z(f). f\∈A Just as in §., by the same proofs, we have

Z(A)=Z(√(A)), and these sets are the closed sets of a topology on Spec(R). In particular, to establish Z(a) Z(b)=Z(a b) ∪ ∩ corresponding to ( ) on page , we need that the functions p fp on Spec(R) take values‡ in integral domains; and this is the case,7→ since fp R/p. ∈ The inverse image of the basic closed set Z(f) under x I (x) is just 7→ K ZL(f), since

x Ln : I (x) Z(f) = x Ln : f I (x) { ∈ K ∈ } { ∈ ∈ K } = x Ln : f(x) = 0 { ∈ } = ZL(f).

n Thus the function x IK (x) from L to Spec(R) is continuous, and moreover, every closed7→ subset of Ln is the inverse image of a closed subset of Spec(R). The argument here uses nothing about L (except L K, as always). ⊇ The function x I (x) injective on Kn, since if a Kn then 7→ K ∈ I (a)=(X0 a0,...,Xn−1 an−1). K − − The map is not generally injective: if n = 2, K = Q, and L is R or C, then I ((π, π))=(X Y )=I ((e, e)). K − K The map is not generally surjective, even if L is large enough to ensure alg a = IK (ZL(a)) for radical a: If L = Q , then (X Y ) is not in the range. But still the map is surjective when L is large− enough, not just ‘algebraically’, but also ‘transcendentally’:

 . The spectrum of a polynomial ring

alg Lemma. If K(X) L, then every prime ideal of K[X] is IK (a) for some a in Ln. ⊆

Proof. Suppose p is a prime ideal of K[X]. We can embed K[X]/p in L over K. Let ai be the image of Xi + p under this embedding. Then for all f in K[X] we have that f(a) is the image of f + p. Therefore IK (a)= p.

alg n Theorem . If K(X) L, then the map x IK (x) from L to Spec(R) is surjective, continuous,⊆ closed, and open.7→

Proof. The lemma gives surjectivity. Before that, we proved continuity. We also proved that closed subsets of Ln are inverse images of closed subsets of Spec(R). Taking complements, we have that open sets are inverse images of open sets. By surjectivity, a set is the inverse image only of its own image; so x I (x) must be closed and open. 7→ K

Thus, if K(X)alg, then Ln is nearly indistinguishable from Spec(R). However, as we observed with (π, π) and (e, e), the former space can con- tain distinct, but topologically indistinguishable, points: distinct points a and b such that every open set containing one of them contains the other. This just means IK (a)=IK (b), that is, the two points are sent to the same point of Spec(R). The Spec(R) itself is not Hausdorff, but it is ‘T0’: if p and q are distinct points of Spec(R), then there is f belonging to p △ q, and this means Z(f) contains only one of the points. It will be useful to have a notation for the open subsets of Spec(R). If f R, let us write ∈ U = Z(f)c = p Spec(R): f / p . f { ∈ ∈ } If A R, we let ⊆ U = Z(A)c = U = p Spec(R): A p . A f { ∈ 6⊆ } f[∈A

These are the open subsets of Spec(R), and each of them is Ua for some radical ideal a of R.

  Schemes

. Regular functions

At the beginning of the last section, we considered the equation f(x) = 0, where f K[X] and x Ln. We have generally f(x) L, that is, f is a function∈ from Ln to L.∈ There can be other such functions.∈ An arbitrary function h from a subset S of Ln to L is regular (or more precisely K-regular) at a point a of S if there is a neighborhood U of a (in the Zariski topology over K, restricted to S) and there are elements f and g of K[X] such that, for all x in U, f(x) h(x)= . g(x) The function is regular, simply, if it is regular at all points of its do- main. The only regular functions on Ln itself are the elements of K[X]. However, let S = Z (Y 2 X3) r Z (X), S = Z (Y 2 X3) r Z (Y ). 0 L − L 1 L − L These are open subsets of their union. On S0 and S1 respectively there are regular functions h0 and h1 given by y x h (x,y)= , h (x,y)= . 0 x2 1 y These two functions agree on S S , since y2 = x3 for all (x,y) in that 0 ∩ 1 set (and even in S0 S1). Thus h0 h1 is a regular function h on S0 S1. However, there are∪ no f and g in∪ K[X,Y ] such that, for all (x,y∪) in S S , h(x,y)= f(x,y)/g(x,y). 0 ∪ 1 2 3 In the example, S0 S1 is an open subset of the closed subset ZL(Y X ) of L2. For now, we∪ shall look just at open subsets of the powers− Ln themselves. If p is a prime ideal of K[X], and f and g in K[X] are such that x n 7→ f(x)/g(x) is well-defined (and therefore regular) on L r ZL(p), this means f/g is a well-defined element of the local ring K[X]p. Now write R = K[X] as before, and let a be an arbitrary radical ideal of R, so that Ua is an open subset of Spec(R). We define shall define a sub-ring, to be denoted by O(Ua),

 . Regular functions of the product

Rp. p Y∈Ua See Figure .. Elements of this product are functions on Ua; so as

Rp

b p Spec(R)

Figure .: A stalk of a sheaf (see p. ) before we have a notion of being regular: An element h of the product is regular at a point p of Ua if, for some open subset V of Ua that contains p, there are f and g in R such that, for all q in V ,

f h = . q g

Note that this requires g / q. The ring O(Ua) consists of the elements of R that are regular∈ at all points of U . p∈Ua p a QThere is a simpler definition when a is a principal ideal (g). In this case, one shows O(U ) = gk : k ω −1R, (g) ∼ { ∈ } n n because the map x/g (x/g : p U(g)) from this ring to O(U(g)) is injective and surjective.7→ See Hartshorne∈ [, Prop. II.., p. ].

 Note well that the factors of the product are the localizations Rp, rather than, say, the quotient-fields of the quotients R/p. However, in the other case that we shall be interested in, where R is itself a product of fields, then the integral domains R/p will already be fields, which are isomorphic to the localizations Rp. See §. below.

  Schemes

If U and V are open subsets of R such that U V , then the restriction- ⊇ U O map from p∈U Rp to p∈V Rp itself restricts to a map ρV from (U) to O(V ). If h O(U), we then write Q ∈ Q U ρV (h)= h ↾ V.

The function U O(U) on the collection of open subsets of R, to- 7→ gether with these homomorphisms ρUV , is called a pre-sheaf of rings on Spec(R) because:

O(∅)= 0 , ρU = id , ρU = ρV ρU . { } U U W W ◦ V U (The notation ρV implies U V ; so for the last equation we have U V W .) We now have a situation⊇ that is ‘dual’ (because the arrows are⊇ ⊇ reversed) to that of the Galois group Gal(Fp): see page . For all p in Spec(R), Rp has a certain ‘universal property’ with respect to the system of rings O(U) such that p U: ∈ U O . Rp is a ring A to which there is a homomorphism hA from (U) for such that, if U V , then ⊇ hV ρU = hU . A ◦ V A . For every such ring A, there is a unique homomorphism h to A from Rp such that hU = h hU . A ◦ Rp See Figure .. Therefore Rp is called a co-limit or direct limit of the given system of rings. This limit can be obtained as a quotient of O the sum p∈U (U) by the smallest ideal that contains, for each pair U and V such that U V , every element x such that x = ρU (x ), and P ⊃ V V U xW = 0 if W is not U or V . The pre-sheaf U O(U) is further a sheaf of rings because it has two additional properties:7→ . If h O(U), and h ↾ V = 0 for all V in an open covering of U, then h =∈ 0. . If there is hV in O(V ) for every V in an open covering of U, and

h ↾ (V W )= h ↾ (V W ) V ∩ W ∩

 . Generic points and irreducibility

Rp E Y✸ ☛☛ ✸✸ ☛☛ h ✸✸ V ☛ ✸ U hR ☛  ✸ hR p ☛☛ A ✸✸ p ☛ ②< b❊❊ ✸ ☛☛ ②② ❊❊ ✸✸ ☛ ②② ❊❊ ✸ ☛ ② V U ❊ ✸ ② h h ❊ ✸ ☛☛②② A A ❊ O O (V ) o U (U) ρV

Figure .: The universal property of Rp

for all V and W in this open covering, then for some h in O(U), for each V in the open covering,

hV = h ↾ V.

The local ring Rp is the stalk of the sheaf at p. In the fullest sense, the spectrum of R is Spec(R) as a topological space equipped with this sheaf. The sheaf is then the structure sheaf of the spectrum of R.

. Generic points and irreducibility

This section is here for completeness, but will not be used later. Every n point a of L is called a generic point of ZL(IK (a)); more precisely, a is a generic point over K of ZL(IK (a)). In the example on page , (π, π) and (e, e) are generic points of Z (X Y ) over Q. L − In any case, if for some radical ideal a, the algebraic set ZL(a) has a generic point, then a must be prime. The converse may fail. For example, alg ZL((X Y )) has no generic point if L Q . However, to Theorem , we have− the following ⊆ Corollary. If K(X)alg L, then the zero-locus in L of every prime ideal has a generic point.⊆

A closed subset of Ln is called irreducible if it cannot be written as the union of two closed subsets, neither of which includes the other.

  Schemes

Theorem . For all radical ideals a of K[X], if Kalg L, ⊆ a is prime Z (a) is irreducible. ⇐⇒ L

Proof. If p is prime, and ZL(p) = ZL(a) ZL(b) for some radical ideals a and b, then (by Hilbert’s Nullstellensatz)∪

p = a b, ∩ so we may assume p = a and therefore Z (a) Z (b). L ⊇ L Suppose conversely Z (a) is irreducible, and fg a. Then L ∈ Z (a) = Z (a f ) Z (a g ), L L ∪ { } ∪ L ∪ { } so we may assume ZL(a) = ZL(a f ) and therefore (again by Hilbert’s Nullstellensatz) f a. ∪ { } ∈ For example, Ln itself is irreducible, since the zero-ideal of K[X] is prime. Therefore the closure of every open subset is the whole space Ln. In any case, every closed set is the union of only finitely many irreducible closed sets: this is by the corollary to the Hilbert Basis Theorem (Theorem  on page ). Hence every radical ideal of K[X] is the intersection of just finitely many elements of Spec(K[X]).

. Affine schemes

The construction of the spectrum of K[X] can be carried out for any ring (that is, commutative unital ring). For an arbitrary ring R, we can still let Spec(R) be the set of prime ideals of R. Then every element f of R determines a function p fp on Spec(R), where fp is the element f + p of R/p. However, the corresponding7→ map

f (fp : p Spec(R)) ( ) 7→ ∈ ∗ from R to p∈Spec(R) R/p need not be injective. For example, the kernel of this map contains X +(X2) when R = K[X]/(X2). Q We do have the generalization of Theorem  on page :

 . Affine schemes

Theorem . For all subsets A of R,

√(A)= p Spec(R): A p . { ∈ ⊆ } \ Proof. It is clear that the intersection includes √(A). Now suppose x ∈ Rr√(A). Let b be an ideal of R that is maximal with respect to including √(A), but not containing any power of x. Say y and z are not in b. By maximality,

x b +(y), x b +(z), ∈ ∈ and therefore x2 b +(yz), ∈ so yz / b (since x2 / b). Thus b is prime. ∈ ∈ Corollary. Spec(R)= √ 0 . { } T In particular, the kernel of f (fp : p Spec(R)) is √ 0 . (Again this is non-trivial if, for example, R7→= K[X]∈/(X2).) { } As before, we still obtain a Galois correspondence between certain subsets of R and of Spec(R). If A R and B Spec(R), we define ⊆ ⊆

Z(A)= p Spec(R): fp = 0 = p Spec(R): A p , { ∈ } { ∈ ⊆ } f\∈A I(B)= f R: fp = 0 = B. { ∈ } p \∈B \ As before, Z(A)=Z(√(A)). So the sets Z(A) (where A R) are just the sets Z(a) (where a is a radical ideal of R). Moreover,⊆ by the last theorem, if a and b are distinct radical ideals, then Z(a) = Z(b). Thus we have: 6 Theorem  (Nullstellensatz). The functions V I(V ) and a Z(a) establish a one-to-one order-reversing correspondence7→ between the7→ radical ideals of R and certain subsets of Spec(R).

  Schemes

If A R, we define ⊆ U = Z(A)c = p Spec(R): A p . A { ∈ 6⊆ }

Theorem . The subsets Ua of Spec(R) are the open sets of a topology on Spec(R).

Proof. Since the elements p of Spec(R) are prime, we have

ab p a p or b p ⊆ ⇐⇒ ⊆ ⊆ or rather ab * p a * p & b * p. ⇐⇒ Therefore Ua Ub = Uab . ∩ Also, Spec(R) = U(1). Finally, if is a collection of ideals of Spec(R), then A Ua = UP A . a [∈A (As a special case, ∅ = U{0}.)

The topology just given is of course the Zariski topology. Just as before, we obtain the sheaf U O(U) of rings on Spec(R), with stalks 7→ Rp. Continuing the example on page , we may let

R = K[X,Y ]/(Y 2 X3). − Let x and y be the images of X and Y respectively in R. Then

U = U U , (x,y) x ∪ y and y x p U = Rp, p U = Rp, ∈ x ⇒ x2 ∈ ∈ y ⇒ y ∈

2 and if p U U , then y/x and x/y are the same element of Rp. Thus ∈ x ∩ y we obtain an element of O(U(x,y)).

 . Regular rings (in the sense of von Neumann)

The spectrum of a ring is called an affine scheme. One point of intro- ducing this terminology is that a scheme, simply, is a topological space with a sheaf of rings such that such that every point of the space has a neighborhood that, with the restriction of the sheaf to it, is an affine scheme. However, we shall not look at schemes in general. In fact we shall look at just one affine scheme whose underlying ring is not a polynomial ring.

. Regular rings (in the sense of von Neumann)

Again R is an arbitrary ring (commutive and unital as always), and p is a prime ideal. We have worked with both the quotient R/p and the localization Rp. Are these two rings ever isomorphic? More precisely, is there ever a well-defined isomorphism x x + p 7→ 1 from R/p to Rp? There is, if R is already a field. • There is not, if R is an integral domain that is not a field; for in • this case the homomorphism from R to Rp is a proper embedding. The homomorphism x + p x/1 is well-defined if and only if 7→ x x p = = 0, ∈ ⇒ 1 that is, if x p, then x s = 0 for some s in R r p. In particular, we must have p I ∈(the set 0· zero-divisors , defined on page ). ⊆ 0 { } ∪ { } The homomorphism x + p x/1 is injective if and only if 7→ x =0 = x p, 1 ⇒ ∈ that is, if x s = 0 for some s in R r p, then x p. Since p contains 0 and is prime,· the homomorphism is automatically∈ injective (if it is well-defined).

  Schemes

The homomorphism is surjective if and only if, for all x/y in Rp, there is z in R and there is s in R r p such that (x yz) s = 0, that is, − · xs = yzs.

It is enough if, for some z, xy = yzy. It is then enough if z has the form xy∗ for some y∗. Then it is enough if y = yy∗y. A ring in which for every element y there is y∗ with this property is called a regular ring (in the sense of von Neumann).

A Boolean ring is a regular ring: just let x∗ be 1 or x (or anything in between). We have shown in effect that every such ring embeds in a Ω power F2 of the two-element field. More generally, if (Ki : i Ω) is an indexed family of arbitrary fields, then the product ∈

Ki iY∈Ω is a regular ring. Indeed, if x R, we can define x∗ by ∈ −1 ∗ xi , if xi = 0, x i = 6 (0, if xi = 0.

Then indeed xx∗x = x. We shall see presently that all regular rings embed in such products.

Theorem . Let R be a regular ring, and let p be a prime ideal, then the map x x + p 7→ 1 is a well-defined isomorphism from R/p to Rp. Moreover, each of these rings is a field. Thus p is maximal.

The definition applies to non-commutative rings as well.

 . The ultraproduct scheme

Proof. For all x in R, we have x = xx∗x, that is,

x (1 xx∗) = 0. · − Since p is a proper ideal, we have

x p 1 xx∗ R r p. ∈ ⇐⇒ − ∈ Thus, as above, the homomorphism is well-defined. It is then injective and surjective, as we said. In particular, in Rp, x 1 = 0 x R r p is well-defined; 1 6 ⇐⇒ ∈ ⇐⇒ x so Rp is a field.

The map in ( ) on page  is now an embedding into a product of fields: ∗

Corollary. Every regular ring R embeds in the product

R/p p∈Spec(Y R) (which is a product of fields) under the map

f (f + p: p Spec(R)). 7→ ∈ Proof. We need only note that the given map is injective, which it is because all ideals of R are radical, so that

Spec(R)= √ 0 = 0 . { } { } \ . The ultraproduct scheme

Now let R be the product i∈Ω Ki of fields as above. As p ranges over Spec(R), the quotients R/p are just the possible ultraproducts of the Q fields Ki. We want to investigate how these arise from the structure

  Schemes sheaf of the spectrum of R. So, letting a be an ideal of R, we want to understand O(Ua). We can identify Spec(R) with Spec(P(Ω)), and more generally, we can P identify ideals of R with ideals of (Ω). Because Rp ∼= R/p, we may assume O(Ua) R/p. ⊆ p Y∈Ua Here we may treat a as an ideal of P(Ω), so Ua can be thought of as an open subset of Spec(P(Ω)). Then, in the product R/p, the p∈Ua index p ranges over this open subset, but in the quotient R/p, the index returns to being the corresponding ideal of R. Q

Let s R/p. Every principal ideal in Ua is (Ω r i ) for some i ∈ p∈Ua { } in Ω. In this case we have a * (Ω r i ), that is, Q { } i a. ∈ [ Let us denote (Ω r i ) by p(i). There is only one prime ideal of P(Ω) that does not contain{ } i , namely p(i). Thus { } U = p(i) . {i} { } In particular, s is automatically regular at p(i). We want to understand when s is regular at not-principal ideals. Still considering also the principal ideals, we have

R/p(i) ∼= Ki.

Let sp(i) be sent to si under this isomorphism, so whenever x in R is such that xi = si, we have sp(i) = x + p(i).

By definition, we have s O(Ua) if and only if, for all p in Ua, for some ∈ subset Ub of a such that b * p, for some x in R, for all q in Ub,

sq = x + q.

We may assume b is a principal ideal (A), where A a r p. If q in UA here is the principal ideal p(j), so that j A, we must∈ have x = s . ∈ j j

 . The ultraproduct scheme

More generally, q UA means A / q, so A is q-large, and hence for all x in R, x + q is determined∈ by (x :∈i A). Thus we may assume i ∈ x =(s : i Ω). i ∈ This establishes that O(U ) is the image of R in R/p: a p∈Ua

O(Ua)= (x + p: p Ua): x QR . { ∈ ∈ }

In particular, O(Ua) is a quotient of R, that is, a reduced product of the Ki. More precisely, O (Ua) ∼= R/b, where b = p = p. p ∈\Ua a\*p It follows that O(Ua) = K . ( ) ∼ i † i∈YS a We can see this in two ways. For example, if p Ua, so that a * p, ∈ then a / p, that is, a is p-large. Therefore the image of x in O(Ua) ∈ depends only on (xi : i a). This shows that O(Ua) is a quotient of S S∈ Ki. i∈S a S ItQ is moreover the quotient by the trivial ideal. For, if i a, then ∈ p(i) Ua, so that x + p(i) depends only on xi, that is, ∈ S x + p(i) = 0 x = 0. ⇐⇒ i This gives us ( ). † Note that possibly a / p, although a p. Such is the case when p is non-principal, but a is the∈ ideal of finite sets.⊆ However, we always have S a / p = ( a) * p. ∈ ⇒ [ [ Another way to establish ( ) is to show † p = (Ω r a). a\*p [

  Schemes

If X Ω r a, and a * p, then a / p, so Ω r a p, and therefore ⊆ ∈ ∈ X p. Inversely, if X * Ω r a, then X a has an element i, so that ∈ ∩ X / p(i) andSa * p(i). S S ∈ S S Because the stalk Rp is always a direct limit of those O(U) such that p ∈ U, we have in the present situation that the ultraproduct i∈Ω Ki/p is a direct limit of those products Ki such that A / p. Symbolically, i∈A ∈ Q Q Ki/p = lim Ki : A / p . −→ ∈ iY∈Ω niY∈A o Equivalently, the ultraproduct is the direct limit of those R/a such that a is a principal ideal included in p:

Ki/p = lim Ki/(B): B p . −→ ∈ iY∈Ω niY∈Ω o

 Bibliography

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 Bibliography

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[] Robin Hartshorne, Algebraic geometry, Springer-Verlag, New York, , Graduate Texts in Mathematics, No. . MR MR ( #)

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